# Copyright 2011 Casa Software Ltd. Centre of Mass

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of centreof-mass is one such mathematical device for reducing the complexity of a problem to a more tractable system for which an understanding can be attempted in terms of mathematics. The question of why loading a set of shelves from the top is more likely to induce an accident than if loaded first from the bottom is one such problem addressed by reducing the shelves and boxes to a single mass at a location through which the stability of the shelves can be understood. The centre-of-mass for a rigid body is central to almost all the solutions so far seen in this text. Each time a ladder is represented by a uniform rod, or a cricket ball is modelled as a particle, the essential idea behind centre-of-mass is deployed, namely, there exists a point in space through which the weight of these bodies acts. The physical dimensions are then only important in terms of the turning effect resulting from the rigid nature of the bodies. While the centre of mass can be determined for 3D objects, the subject will be developed only for 1D and 2D rigid bodies. A 2D rigid body is referred to in mechanics as a lamina, which is characterised by a body with some mass and having an appreciable plane area, but negligible thickness. Discussions will be further limited to uniform lamina and networks of uniform rods. The significance of the term uniform is the centre of mass for a uniform rigid body can be determined based on the geometric lines of symmetry. Since the centre of mass for a uniform lamina must lie on a line of geometric symmetry, two or more geometric lines of symmetry cross at the position for the centre of mass. Thus, a uniform lamina with the geometry of a circle will have the centre of mass coinciding with the geometric centre for the circular shape. Provided a complex lamina can be broken down into a set of shapes for which the centre of mass is known, the centre of mass for complex shaped lamina can be determined from the techniques described below. 1

2 Calculating the centre of mass is performed by replacing the uniform lamina by a light lamina for which a single particle of mass equal to the mass of the uniform lamina is attached to the light lamina such that the turning effect under the influence of gravity for the two laminas about any line within the plane of these laminas is the same. For a square uniform lamina lying in the horizontal plane, the turning effect of the lamina would be the same as that of a particle positioned at the point where two axes of symmetry cross, with the same mass as the uniform lamina. Uniform lamina mass M Light lamina with particle mass M at centre-ofmass for the uniform lamina When discussing turning effects in 2D, the significant statement is the turning effect is now about a line rather than a point. For 1D problems, where an object is modelled using a rigid rod, the turning effect was considered to be about a point. Another way of thinking about the 1D concept of rotation about a point is that the point is the cross-section of a line perpendicular to the rod and the plane of the paper on which the rod is drawn. In this sense moments, even for 1D problems, were always about a line, not a point. For more complicated shaped laminas, the problem of determining the centre of mass is that of reducing the lamina to shapes for which centre of masses are known, and then determining the centre of mass for a set of particles with masses and positions determined by these simpler shapes. For example, a square lamina with a quadrant missing could be broken down into a rectangle and a square. The centre of mass for the rectangle and square are determined by geometric symmetry, and these component lamina replaced by particles of equivalent masses. 1M 3M 2M The centre of mass for the original shape is then determined from the centre of mass for a set of particles of different masses attached to a light lamina. 2

3 The light lamina to which these particles are attached is only conceptual and can be viewed as any shape we wish, which is important for situations where the centre of mass lies outside the boundary of the original lamina. The centre of mass for a hollow object, such as a washer, is not within the material part of the washer. If a mechanical system can be reduced to a set of particles of known mass and spatial separation, then the centre of mass for the original rigid body can be calculated by summing the moments for each particle about any line we choose, and then calculating the distance from the line for a particle equal in mass to the total mass such that the moment of the particle of total mass is the same as the sum of the moments for all the particles. The sum of moments must account for a specified rotational sense with respect to the chosen line. For the component particles, moments about the y-axis: Must be equivalent to the moment for the total mass particle: where is the distance of the centre of mass from the y-axis. Therefore, Moments about the x-axis for component particles: Must be equivalent to the moment for the total mass particle where is the distance of the centre of mass from the x- axis. Therefore, 3

4 3M Centre of Mass In general, provided a lamina can be reduced to a set of particles of known mass and positions, applying the same logic for 2D shapes based on moments about the coordinate axes chosen for the shape, the centre of mass can be determined as a weighted average of position: If the total mass, Similarly, Note, depending on where the coordinate axes are chosen for a particular shape, the particle coordinates may be positive or negative. These positive and negative coordinate values correspond to the convention used for the rotational sense of the moments underlying these formulae. 4

5 Example A uniform plane lamina constructed from a rectangle connected to a semi-circular section and an isosceles triangle with dimensions,, and. Show that the centre of mass for the lamina is below the line. The lamina is constructed from three components: the rectangle, a semi-circle and a triangle connected to the rectangle at and. The first step is to calculate the centre of mass for a semicircle using calculus. The triangle also needs further analysis, but the centre of mass for the rectangle is determined from geometric symmetry. Centre of Mass for a Semi-circle Equation of a circle Mass of strip with area density is By geometric symmetry, the centre of mass must lie on the x-axis for a semi-circle radius the functional form for. obeying 5

6 Calculating distance for the centre of mass from the y-axis for a semi-circle of uniform density per unit area is performed using the concepts of summing moments in exactly the same way moments are calculated for sets of particles. The semi-circle is approximated by a set of rectangular regions of width and height. By symmetry, the centre of mass for these rectangles lies on the x-axis and the mass for each rectangle positioned from the y-axis is Where. The total mass for the semi-circle is known exactly, and is mass for the semi-circle at the, yet to be determined, position acting through the centre of from the y-axis. If the centre of mass for the semi-circle is mass will be from the y-axis, then an approximation to the centre of In the digital age, such an expression would be sufficient to obtain an answer within the precision achieved by a calculator or a computer, however, calculus can be used to derive an exact expression for the centre of mass. If the number of rectangles used in this approximation is allowed to increase, the width for each rectangle decreases, thus as rectangles are added to the approximation, the area and therefore the mass for each rectangle gets smaller. By increasing the number of rectangles, the number of small masses becomes ever larger, and understanding the consequences of adding more and more of ever 6

7 smaller items is precisely the stuff of calculus. Mechanics, differentiation and integration are strongly linked. To obtain the centre of mass for the semi-circle, as sum involved in must go to zero as ) the summation tends to the integral (the Using the substitution, with limits from to. This result is a standard result from a host of other standard centres of mass for uniform laminas, all of which will be provided under examination conditions, but nevertheless, illustrates where a desire to understand a physical object leads to the concepts of calculus and therefore should be seen as the motivation for studying techniques of integration in other mathematics courses. Centre of Mass for an Isosceles Triangle Equation of a line Mass of strip with area density is By symmetry, the centre of mass for an isosceles triangle lies on the x-axis and the mass for each rectangle positioned from the y-axis is Where. The total mass for the isosceles triangle is the triangle at the position from the y-axis. acting through the centre of mass for 7

8 If the centre of mass for the triangle is mass will be from the y-axis, then an approximation to the centre of Again moving to the limit as The centre of mass for the complex shape can now be reduced to a set of three particles located at the centres of mass for each of the component parts. To calculate the distance of the centre of mass for the entire lamina from the line, the moment of the weights for these three particles about must be the same as the moment about of a single particle of total mass. Thus, 8

9 Since the centre of mass for the particle corresponding to the rectangle is on the opposite side of the line to the two particles corresponding to the triangle and the semi-circle, the moment for the rectangle is negative relative to the two particles above the line. The value for is negative, therefore the centre of mass is below the line. Example A metal bracket is designed as a square ABCD of size 12 cm with a square of size 4 cm removed from one quadrant as shown in the diagram. a) Find the distance of the centre of mass for the bracket from the side BC. The bracket is suspended from A and hangs at rest. b) Find the size of the angle between AB and the vertical. The problem differs slightly from the previous example by virtue of a missing square of metal rather than shapes being added to a square. The problem could be approached by dividing the metal bracket into many smaller rectangles from which the centre of mass for the bracket could be calculated. However, by applying mathematical reasoning, the problem can be solved by considering three masses corresponding to: 1. The mass of a square bracket without the removal of the smaller square. 2. The mass of the square corresponding to the missing smaller square. 3. The mass of the bracket with the smaller square removed. The reasoning is as follows. The bracket before the smaller square is punched out consists of the union of the bracket after the smaller square is removed and the small square itself. If the metal plate from which the bracket is made is considered to be constructed from the bracket plus the smaller square, the centre of mass for the square metal plate can be expressed in terms of the two component parts. The only difference is the unknown is now one of the component parts. 9

10 Since the centre of mass for the larger square is known and the centre of mass for the smaller square is also known, both obtained using the lines of symmetry for a square, the unknown centre of mass for the bracket can be determined. a) Find the distance of the centre of mass for the bracket from the side BC. Assuming the metal bracket can be modelled as a uniform lamina of density per unit area, the masses for a square of size by, the smaller square of size by and the bracket are calculated as follows. The same logic therefore applies as before, namely, the moments about the x-axis for the full square must be the same as the moments about the x-axis for the two component parts. By symmetry,. 10

11 b) Find the size of the angle between AB and the vertical. The lamina is suspended from the corner and is assumed to be at rest. The practical method for determining the centre of mass for a metal bracket is to suspend the bracket from two different corners and each time mark the line on the bracket through the corner from which the bracket is suspended and a plumb line hanging vertically downwards. Two such lines marked on the metal bracket would intersect at the centre of mass. If the bracket is suspended at rest from the corner, then the mathematical solution is therefore obtained by drawing a line through which passes through the centre of mass previously calculated. Vertical line 11

12 Example Three particles are attached to a uniform rectangular lamina. The coordinates for the particles are, and corresponding to masses, and, respectively. a) Given that the centre of mass for these three particles in the absence of the uniform lamina has x-coordinate equal to, calculate the value of. b) If the mass of the uniform lamina is, find the coordinates for the centre of mass of the combined system, namely, lamina and three attached masses. c) If the combined system is freely suspended from the corner, calculate the angle between AB and the horizontal. Solution a) Given that the centre of mass for these three particles in the absence of the uniform lamina has x-coordinate equal to, calculate the value of. By taking moments about the y-axis, the particle of mass will be eliminated from the calculation while the particle specified as will be part of the equation. Thus, taking moments about the y- axis for the three particles in the absence of the lamina, b) If the mass of the uniform lamina is, find the coordinates for the centre of mass of the combined system of lamina and three masses. Since the first part of the question gave the x-coordinate for the centre of mass for the three particles attached to a light lamina, calculating the corresponding y-coordinate for the centre of mass offers a means of reducing the problem from a four mass to a two mass problem. The three masses attached to the lamina will be reduced to a single particle of mass positioned at the calculated centre of mass for these three individual masses. The y-coordinate for the three particles of mass, and is obtained by taking moments about the x-axis. 12

13 The centre of mass for the three particles is therefore. A particle of mass 16m positioned at has the same moment as the three separate particles. The centre of mass for the combined system of uniform lamina mass and the three particles is obtained by finding the centre of mass for the two particle of mass positioned at and positioned at. Taking moments about the y-axis: Taking moments about the x-axis: c) If the combined system is freely suspended from the corner, calculate the angle between AB and the horizontal. Suspending the combined system from means the centre of mass will be vertically below, therefore the angle between and the horizontal is the angle marked on the diagram. Horizontal line Toppling Points for a Lamina If the reaction force to a lamina placed on a plane is modelled as a force acting perpendicular to the plane through the point vertically below the centre of mass for the lamina, the problem of stability 13

14 with respect to toppling can be understood in terms of turning effects about the point vertically below the centre of mass. Placing a lamina on an inclined plane causes the position for the reaction force to change. For a rough surface a friction force must also act through the point vertically below the centre of mass. If all the forces acting on a body act through the same point, since the distance from that point to the line of action for all the forces is zero, the moment about that point is zero. For a lamina in contact with a plane, the reaction and friction forces can only act through a point of the lamina in contact with the plane, thus provided the centre of mass is vertically above any point in contact with the plane, the sum of the moments for all the forces must be zero and therefore the body modelled by the lamina will be stable with respect to toppling. Tipping point: the centre of mass is vertically above the corner for the edge in contact with the plane. Stable: The centre of mass is vertically above a point within the base. Tipping occurs: It is not possible for all forces to act through a single point. 14

15 Example A uniform lamina is constructed from an isosceles triangle with base connected to a rectangle of size by as shown in the diagram. and height a) Determine the distance of the centre of mass from the common edge between the triangle and the rectangle. The lamina is placed on an inclined plane which makes an angle to the horizontal. b) Determine the largest angle for an inclined plane such that the lamina does not topple. The solution for part a) is required in the solution of part b). The angle at which the lamina is at the point of tipping occurs when the centre of mass is vertically above the point on the base of the lamina in contact with the plane about which the lamina will rotate when toppling. a) Determine the distance of the centre of mass from the common edge between the triangle and the rectangle. Let the distance of the centre of mass for the lamina from the line joining the triangle to the rectangle be. Since the lamina has a line of symmetry passing through the mid-point of the triangle base and the mid-point of the rectangle edge parallel to the triangle base, the centre of mass for a uniform lamina must lie on this line of symmetry. The coordinate axes can be positioned with respect to the lamina as follows. 15

16 The y coordinate of centre-of-mass for the triangle, the rectangle and the combined lamina all have the same value, namely,. The distance lamina. is then calculated to be the same as the x coordinate for the centre of mass for the The centre of mass for the lamina is calculated using prior knowledge about the centre of mass of a triangular uniform lamina and the centre of mass for a rectangular uniform lamina. If the mass per unit area is, then the information used in calculating the centre of mass for the combined lamina is as follows. The centre of mass for any triangle, not just an isosceles triangle, lies on the median line passing through a vertex and the mid-point of the opposite side to the triangle. It can be shown that the distance from the mid-point of a side to the centre of mass is a third the length of the median line passing through the opposite vertex. The distance from the y-axis for the triangular lamina of height 6 cm is therefore to the left of the y-axis so the x-coordinate for the centre of mass is. 16

17 Median lines drawn from the mid-point of a side to the opposite vertex intersect at the centre of mass for a scalene triangle. Taking moments about the y-axis, b) Determine the largest angle for an inclined plane such that the lamina does not topple. The tipping point for the lamina placed on an inclined plane is determined by assuming the centre of mass is vertically above the corner about which rotation can occur. 17

18 Vertical Line The angle for the inclined plane is therefore given by 18

21 Thus, the magnitude of the friction force is given by b) Determine the angle made by the ladder with the horizontal floor. The angle is obtained by generating equations involving lengths and forces. For a rigid body to be in equilibrium, the moments of the forces taken about any point must sum to zero. Since the condition for equilibrium must be satisfied for any point, the best point to choose is a point on the ladder which simplifies the resulting equations. Taking moments about eliminates the normal reaction force and the friction force from the equations. The moment of a force about a point is the product of the magnitude of the force and the perpendicular distance from the point to the line of action of the force. F For the ladder, the distance from the point to the line of action of the various forces will introduce the angle. Further, since the line of action for both the friction and the normal reaction force at both pass through, the distance between and these two forces is zero, hence the earlier statement asserting the friction and normal reaction forces at are eliminated by choosing as the point about which moments are taken. Moments about, taking clockwise direction as positive: 21

22 Moment about Copyright 2011 Casa Software Ltd. for Normal Reaction Force at Wall Moment about for Weight of particle at Moment about for Weight of Ladder Therefore, summing the moments about yields: This equation involves two unknowns, therefore another equation is required involving. Resolving the forces in the horizontal direction. 22

24 Ladder of length of Mass The ladder and masses can be placed on the horizontal ground for the convenience of the calculation. A similar more involved calculation could be performed with the ladder against the wall, but the outcome for the centre of mass must be the same, which will be shown by solving the problem as stated by reducing the system of masses to a single effective mass and by direct solution treating each mass individually. Placing the ladder and masses on the horizontal floor, the moment of a force is the magnitude of the force times the perpendicular distance to the line of action of the force. All weights act vertically, hence the desire to lay the ladder flat. Taking moments about the point A for each of these two equivalent mechanical systems yields: For the heavy rod and masses: For the light rod and single mass: If these two systems are equivalent the moments about A must be equal. Thus, Those studying statistics will recognise the quotient as a weighted mean where the mass is equivalent to the frequency. 24

25 The original problem is then reduced to a light rod and a single particle: The solution to the original problem then proceeds by listing the equations generated by resolving the forces vertically and horizontally as indicated on the diagram, the relationship between friction force and the normal reaction force of the ladder on the floor and taking moments about A. Resolving vertically: Resolving horizontally: Finally taking moments about A: Coefficient of friction: Equation (1) yields, therefore Equation (2) shows that and substituting for and Equation (3) gives 25

26 Since, Copyright 2011 Casa Software Ltd. Substituting into Equation (5) provides the inequality For equilibrium, the ladder must have a mass placed at A where,. The alternative to reducing the ladder and masses to a single mass is the direct approach where all the forces for each explicit particle and the ladder are detailed on the diagram. The solution proceeds with exactly the same steps, the difference being the complexity of the equations generated from resolving vertically and taking moments about the point. Taking a complex system and reducing the system to an equivalent simpler system is a common theme in higher mathematics, and thinking in terms of the centre of mass is a good example illustrating this approach. It also explains why calculating the centre of mass for complex rigid bodies features in most mechanics courses. Resolving vertically: Moments about A: These equations should be compared to Equations (1) and (3) above. The result obtained from these equations is the same as when the ladder and particles are reduced to a single mass acting at the centre of mass for the system. 26

27 Example A uniform rod of length and mass rests in equilibrium with one end in contact with a rough horizontal surface at, and is supported by a smooth peg at position where. When in limiting equilibrium, the rod makes an angle with the horizontal surface. Show that a) The normal reaction force at is. b) The coefficient of friction can be expressed in terms of the angle by The essential difference between a problem involving a ladder leaning against a wall and a rod supported by a peg is at the contact point for the rod. If the peg and the wall are smooth, that is the contact point between the rod and either the wall or the peg has no friction force therefore only involves a normal reaction force, the difference between these two problems is the direction for the reaction force. The normal reaction force of the ladder in contact with a vertical wall is horizontal, while for the smooth peg, the direction for the normal reaction force depends on the angle between the rod and the floor. Normal reaction force for a ladder against a smooth wall Normal reaction force for a rod supported by a smooth peg If a ladder were placed on top of a smooth wall, the problem reduces to that of a smooth peg. 27

28 Ladder placed on top of a smooth wall Solution The most important part of the solution for the uniform rod supported by a smooth peg is to draw a diagram annotated with all the force vectors. a) The normal reaction force at is. The first part to the question is intended to guide the solution in a direction leading to the answer for part b). The pattern common to solving statics problems involves resolving the forces in two perpendicular directions and taking moments about an appropriate point to obtain equations relating the key information desired. Since the normal reaction force at is the information requested first, resolving the force vectors in the vertical direction and, for equilibrium, equating to zero would be a good first step in the solution. 28

29 Resolving the forces vertically: Resolving the forces Horizontally: Since act through taking moments about : The expression for the normal reaction force at is therefore obtained by substituting Equation (3) into Equation (1) And since 29

30 b) The coefficient of friction can be expressed in terms of the angle. For limiting equilibrium Substituting Equation (2) into Equation (5) Substituting from Equation (3), and from Equation (4) And since, 30

31 Example A uniform rod of length and mass rests with one end on a rough horizontal plane. A force is applied to the rod at a point which is from such that the rod is held in limiting equilibrium at an angle to the horizontal, where. The line of action of the force at is in the same vertical plane as the rod. Given that the coefficient of friction between the rod and the horizontal plane is ground on the rod at., find the magnitude of the normal reaction force of the The force acting at is not specified by a magnitude or direction. Since so little is known about the force at the solution is likely to involve taking moments about, as in so doing the resulting equation for equilibrium of the rod will not require any knowledge about the force at. Further, since the coefficient of friction is given, the friction force is known in terms of the normal reaction at, so taking moments about and substituting into the resulting equation allows the normal reaction to be determined. For equilibrium Moments about Since, and, 31

32 Example A uniform rod of length and mass is smoothly hinged to a vertical wall at. A light rod of length is freely jointed to the rod at and fixed to the wall vertically below at. The rod is positioned with so that the rod is held horizontally in equilibrium. a) Determine the thrust in the rod. b) The magnitude of the force exerted by the wall on the rod at. The problem as stated is aimed at maintaining the uniform rod position. To achieve this aim, all the forces acting on the rod for the rod. in equilibrium in a horizontal must be introduced into a diagram The active force on the rod is the weight of the rod 20g acting through the centre of mass for the uniform rod. Since the rod is uniform, the centre of mass acts through the geometric centre of the rod and is therefore from. Two passive forces, one at the smooth hinge at and the second the thrust of the rod at, must also be included as external forces acting on the rod. These passive forces act in response to the rod s weight, and are only present because of active forces such as the weight of the rod. If the rod were light, no force would exist at the hinge and no thrust would act in the rod. A smooth hinge exerts a force on the rod from the wall with magnitude and direction as yet to be determined. To accommodate the reaction force at, two perpendicular forces are introduced which represent the components of the reaction force at, or in vector notation,, where and are unit vectors in the horizontal and vertical directions. Thrust exerted by the rod is a passive force acting against the force attempting to compress the rod. The thrust acts in a direction parallel to the rod CD and through the point C on the rod AB. 32

33 Solution a) Determine the thrust in the rod. Since the reaction force in the hinge acts through, taking moments about and, for equilibrium, equating to zero provides an equation involving the weight of the rod and the thrust only. The perpendicular distance between the line of action of the weight and the point is. The perpendicular distance between the line of action for the thrust and must be calculated from the geometry of the rods and vertical wall. Moments about A: b) The magnitude of the force exerted by the wall on the rod at. The reaction force at the smooth hinge is defined in terms of two perpendicular components, therefore the magnitude of the reaction force, is. These component forces are obtained by resolving the forces on the rod in the horizontal and vertical directions. 33 and

34 Resolving vertically, Resolving horizontally, The negative sign for the components to the reaction force of the hinge indicates the forces as marked on the diagram are actually in the opposite direction. The magnitude for the reaction force of the hinge at is therefore, Example A particle of weight is attached to the end of a light rod of length. A second particle of weight is attached to the opposite end of the rod. Two light inextensible strings each of length are attached to the ends of the rod before being fixed to a point C, from which the rod is hung. Show that if the rod is in equilibrium, the angle between the rod and the horizontal is where The active forces are due to particles located at and on the light rod. When in equilibrium, these active forces must be balanced by passive forces in the two light inextensible strings. The strings are fixed at and attached at each end of the rod, therefore each string will cause a force to act on the rod due to the strings resistance to being stretched. A force resisting elongation in a string is referred to as tension and is illustrated using an arrow pointing into the string. The solution to the problem of hanging a rod using two strings, as usual, involves drawing a diagram to illustrate the forces acting on the rod and particles, however, the importance of an accurate well 34

35 drawn diagram is clearly shown by this example. Without a clear geometric understanding for the system of strings, particles and rod, the solution is more difficult to see. The problem is posed to extract a higher than usual mathematical content for mechanics questions, but would certainly require an even higher mathematical intuition without a pictorial understanding of the components involved. Geometric points: 1. The triangle is isosceles, with two side of equal length and therefore the angles and are equal angles, say. 2. The angle between and the horizontal. 3. The angle between and the horizontal. The solution involves observing the structure within the mechanical system; specifically the four forces are divided between two points of action, therefore taking moments about each of the two points and yields two equations for the tensions and in terms of the angle. The angle is clearly determined by the isosceles triangle, as follows. Using the geometry for the isosceles triangle:, and. 35

36 Taking moments about and, for equilibrium, equating to zero: Similarly, taking moments about and, for equilibrium, equating to zero: These two equations involve three unknowns,, and, therefore a third equation relating these values is required. Resolving horizontally and, again, using the condition for equilibrium that the resultant force in any direction must be zero, Since and, Using the trigonometric identities And Equations (1) and (2) yield Substituting for into Equation (3), 36

37 Since, 37

### Solving Simultaneous Equations and Matrices

Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering

### Mechanics 2. Revision Notes

Mechanics 2 Revision Notes November 2012 Contents 1 Kinematics 3 Constant acceleration in a vertical plane... 3 Variable acceleration... 5 Using vectors... 6 2 Centres of mass 8 Centre of mass of n particles...

### AN ROINN OIDEACHAIS AGUS EOLAÍOCHTA LEAVING CERTIFICATE EXAMINATION, 2001 APPLIED MATHEMATICS HIGHER LEVEL

M3 AN ROINN OIDEACHAIS AGUS EOLAÍOCHTA LEAVING CERTIFICATE EXAMINATION, 00 APPLIED MATHEMATICS HIGHER LEVEL FRIDAY, JUNE AFTERNOON,.00 to 4.30 Six questions to be answered. All questions carry equal marks.

### Mechanics 1. Revision Notes

Mechanics 1 Revision Notes July 2012 MECHANICS 1... 2 1. Mathematical Models in Mechanics... 2 Assumptions and approximations often used to simplify the mathematics involved:... 2 2. Vectors in Mechanics....

### Chapter 11 Equilibrium

11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of

### 3 rd International Physics Olympiad 1969, Brno, Czechoslovakia

3 rd International Physics Olympiad 1969, Brno, Czechoslovakia Problem 1. Figure 1 shows a mechanical system consisting of three carts A, B and C of masses m 1 = 0.3 kg, m 2 = 0.2 kg and m 3 = 1.5 kg respectively.

### www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x

Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity

### m i: is the mass of each particle

Center of Mass (CM): The center of mass is a point which locates the resultant mass of a system of particles or body. It can be within the object (like a human standing straight) or outside the object

### 1 of 7 4/13/2010 8:05 PM

Chapter 33 Homework Due: 8:00am on Wednesday, April 7, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy [Return to Standard Assignment View] Canceling a Magnetic Field

### AN ROINN OIDEACHAIS AGUS EOLAÍOCHTA LEAVING CERTIFICATE EXAMINATION, 2000

M31 AN ROINN OIDEACHAIS AGUS EOLAÍOCHTA LEAVING CERTIFICATE EXAMINATION, 2000 APPLIED MATHEMATICS - ORDINARY LEVEL FRIDAY, 23 JUNE - AFTERNOON, 2.00 to 4.30 Six questions to be answered. All questions

### EQUILIBRIUM AND ELASTICITY

Chapter 12: EQUILIBRIUM AND ELASTICITY 1 A net torque applied to a rigid object always tends to produce: A linear acceleration B rotational equilibrium C angular acceleration D rotational inertia E none

### 7.3 Volumes Calculus

7. VOLUMES Just like in the last section where we found the area of one arbitrary rectangular strip and used an integral to add up the areas of an infinite number of infinitely thin rectangles, we are

### PSS 27.2 The Electric Field of a Continuous Distribution of Charge

Chapter 27 Solutions PSS 27.2 The Electric Field of a Continuous Distribution of Charge Description: Knight Problem-Solving Strategy 27.2 The Electric Field of a Continuous Distribution of Charge is illustrated.

### SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A lawn roller in the form of a uniform solid cylinder is being pulled horizontally by a horizontal

### Mechanics Lecture Notes. 1 Notes for lectures 12 and 13: Motion in a circle

Mechanics Lecture Notes Notes for lectures 2 and 3: Motion in a circle. Introduction The important result in this lecture concerns the force required to keep a particle moving on a circular path: if the

Week 8 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

### Mathematics 1. Lecture 5. Pattarawit Polpinit

Mathematics 1 Lecture 5 Pattarawit Polpinit Lecture Objective At the end of the lesson, the student is expected to be able to: familiarize with the use of Cartesian Coordinate System. determine the distance

### Announcements. Dry Friction

Announcements Dry Friction Today s Objectives Understand the characteristics of dry friction Draw a FBD including friction Solve problems involving friction Class Activities Applications Characteristics

### Geometry and Measurement

The student will be able to: Geometry and Measurement 1. Demonstrate an understanding of the principles of geometry and measurement and operations using measurements Use the US system of measurement for

### Copyright 2011 Casa Software Ltd. www.casaxps.com

Table of Contents Variable Forces and Differential Equations... 2 Differential Equations... 3 Second Order Linear Differential Equations with Constant Coefficients... 6 Reduction of Differential Equations

### Physics 210 Q ( PHYSICS210BRIDGE ) My Courses Course Settings

1 of 16 9/7/2012 1:10 PM Logged in as Julie Alexander, Instructor Help Log Out Physics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings Course Home Assignments Roster Gradebook Item Library

### Stress and Deformation Analysis. Representing Stresses on a Stress Element. Representing Stresses on a Stress Element con t

Stress and Deformation Analysis Material in this lecture was taken from chapter 3 of Representing Stresses on a Stress Element One main goals of stress analysis is to determine the point within a load-carrying

### Lecture 8 : Coordinate Geometry. The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 20

Lecture 8 : Coordinate Geometry The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 0 distance on the axis and give each point an identity on the corresponding

### 2. THE x-y PLANE 7 C7

2. THE x-y PLANE 2.1. The Real Line When we plot quantities on a graph we can plot not only integer values like 1, 2 and 3 but also fractions, like 3½ or 4¾. In fact we can, in principle, plot any real

### 1 of 7 10/2/2009 1:13 PM

1 of 7 10/2/2009 1:13 PM Chapter 6 Homework Due: 9:00am on Monday, September 28, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]

### Physics-1 Recitation-7

Physics-1 Recitation-7 Rotation of a Rigid Object About a Fixed Axis 1. The angular position of a point on a wheel is described by. a) Determine angular position, angular speed, and angular acceleration

### Physics, Chapter 3: The Equilibrium of a Particle

University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1-1958 Physics, Chapter 3: The Equilibrium of a Particle

### Fric-3. force F k and the equation (4.2) may be used. The sense of F k is opposite

4. FRICTION 4.1 Laws of friction. We know from experience that when two bodies tend to slide on each other a resisting force appears at their surface of contact which opposes their relative motion. The

### Mechanics Cycle 2 Chapter 13+ Chapter 13+ Revisit Torque. Revisit Statics

Chapter 13+ Revisit Torque Revisit: Statics (equilibrium) Torque formula To-Do: Torque due to weight is simple Different forms of the torque formula Cross product Revisit Statics Recall that when nothing

Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced Subsidiary Thursday 12 January 2006 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae (Green or Lilac)

### Section 1.8 Coordinate Geometry

Section 1.8 Coordinate Geometry The Coordinate Plane Just as points on a line can be identified with real numbers to form the coordinate line, points in a plane can be identified with ordered pairs of

### Foundation Engineering Prof. Mahendra Singh Department of Civil Engineering Indian Institute of Technology, Roorkee

Foundation Engineering Prof. Mahendra Singh Department of Civil Engineering Indian Institute of Technology, Roorkee Module - 03 Lecture - 09 Stability of Slopes Welcome back to the classes of on this Stability

### Chapter 12. The Straight Line

302 Chapter 12 (Plane Analytic Geometry) 12.1 Introduction: Analytic- geometry was introduced by Rene Descartes (1596 1650) in his La Geometric published in 1637. Accordingly, after the name of its founder,

### Section 2.1 Rectangular Coordinate Systems

P a g e 1 Section 2.1 Rectangular Coordinate Systems 1. Pythagorean Theorem In a right triangle, the lengths of the sides are related by the equation where a and b are the lengths of the legs and c is

### Physics 1A Lecture 10C

Physics 1A Lecture 10C "If you neglect to recharge a battery, it dies. And if you run full speed ahead without stopping for water, you lose momentum to finish the race. --Oprah Winfrey Static Equilibrium

### 9 ROTATIONAL DYNAMICS

CHAPTER 9 ROTATIONAL DYNAMICS CONCEPTUAL QUESTIONS 1. REASONING AND SOLUTION The magnitude of the torque produced by a force F is given by τ = Fl, where l is the lever arm. When a long pipe is slipped

### Two-Force Members, Three-Force Members, Distributed Loads

Two-Force Members, Three-Force Members, Distributed Loads Two-Force Members - Examples ME 202 2 Two-Force Members Only two forces act on the body. The line of action (LOA) of forces at both A and B must

### EXPONENTS. To the applicant: KEY WORDS AND CONVERTING WORDS TO EQUATIONS

To the applicant: The following information will help you review math that is included in the Paraprofessional written examination for the Conejo Valley Unified School District. The Education Code requires

### Centroid: The point of intersection of the three medians of a triangle. Centroid

Vocabulary Words Acute Triangles: A triangle with all acute angles. Examples 80 50 50 Angle: A figure formed by two noncollinear rays that have a common endpoint and are not opposite rays. Angle Bisector:

### Announcements. Moment of a Force

Announcements Test observations Units Significant figures Position vectors Moment of a Force Today s Objectives Understand and define Moment Determine moments of a force in 2-D and 3-D cases Moment of

### Chapter 5: Distributed Forces; Centroids and Centers of Gravity

CE297-FA09-Ch5 Page 1 Wednesday, October 07, 2009 12:39 PM Chapter 5: Distributed Forces; Centroids and Centers of Gravity What are distributed forces? Forces that act on a body per unit length, area or

### Lecture L22-2D Rigid Body Dynamics: Work and Energy

J. Peraire, S. Widnall 6.07 Dynamics Fall 008 Version.0 Lecture L - D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture L-3 for

### Force on Moving Charges in a Magnetic Field

[ Assignment View ] [ Eðlisfræði 2, vor 2007 27. Magnetic Field and Magnetic Forces Assignment is due at 2:00am on Wednesday, February 28, 2007 Credit for problems submitted late will decrease to 0% after

### Applications of Integration Day 1

Applications of Integration Day 1 Area Under Curves and Between Curves Example 1 Find the area under the curve y = x2 from x = 1 to x = 5. (What does it mean to take a slice?) Example 2 Find the area under

### Eðlisfræði 2, vor 2007

[ Assignment View ] [ Pri Eðlisfræði 2, vor 2007 29a. Electromagnetic Induction Assignment is due at 2:00am on Wednesday, March 7, 2007 Credit for problems submitted late will decrease to 0% after the

### 6. Vectors. 1 2009-2016 Scott Surgent (surgent@asu.edu)

6. Vectors For purposes of applications in calculus and physics, a vector has both a direction and a magnitude (length), and is usually represented as an arrow. The start of the arrow is the vector s foot,

### 4.2 Free Body Diagrams

CE297-FA09-Ch4 Page 1 Friday, September 18, 2009 12:11 AM Chapter 4: Equilibrium of Rigid Bodies A (rigid) body is said to in equilibrium if the vector sum of ALL forces and all their moments taken about

### ENGINEERING DRAWING. UNIT I - Part A

ENGINEERING DRAWING UNIT I - Part A 1. Solid Geometry is the study of graphic representation of solids of --------- dimensions on plane surfaces of ------- dimensions. 2. In the orthographic projection,

### Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

### Biggar High School Mathematics Department. National 5 Learning Intentions & Success Criteria: Assessing My Progress

Biggar High School Mathematics Department National 5 Learning Intentions & Success Criteria: Assessing My Progress Expressions & Formulae Topic Learning Intention Success Criteria I understand this Approximation

### GEOMETRY & INEQUALITIES. (1) State the Triangle Inequality. In addition, give an argument explaining why it should be true.

GEOMETRY & INEQUALITIES LAMC INTERMEDIATE GROUP - 2/09/14 The Triangle Inequality! (1) State the Triangle Inequality. In addition, give an argument explaining why it should be true. Given the three side

### Unit 4: Science and Materials in Construction and the Built Environment. Chapter 14. Understand how Forces act on Structures

Chapter 14 Understand how Forces act on Structures 14.1 Introduction The analysis of structures considered here will be based on a number of fundamental concepts which follow from simple Newtonian mechanics;

### Magnetism Conceptual Questions. Name: Class: Date:

Name: Class: Date: Magnetism 22.1 Conceptual Questions 1) A proton, moving north, enters a magnetic field. Because of this field, the proton curves downward. We may conclude that the magnetic field must

### STRAIGHT LINES. , y 1. tan. and m 2. 1 mm. If we take the acute angle between two lines, then tan θ = = 1. x h x x. x 1. ) (x 2

STRAIGHT LINES Chapter 10 10.1 Overview 10.1.1 Slope of a line If θ is the angle made by a line with positive direction of x-axis in anticlockwise direction, then the value of tan θ is called the slope

### Archdiocese of Washington Catholic Schools Academic Standards Mathematics

5 th GRADE Archdiocese of Washington Catholic Schools Standard 1 - Number Sense Students compute with whole numbers*, decimals, and fractions and understand the relationship among decimals, fractions,

### ascending order decimal denominator descending order Numbers listed from largest to smallest equivalent fraction greater than or equal to SOL 7.

SOL 7.1 ascending order Numbers listed in order from smallest to largest decimal The numbers in the base 10 number system, having one or more places to the right of a decimal point denominator The bottom

### Transformations Packet Geometry

Transformations Packet Geometry 1 Name: Geometry Chapter 14: Transformations Date Due Section Topics Assignment 14.1 14.2 14.3 Preimage Image Isometry Mapping Reflections Note: Notation is different in

### Grade 5 Math Curriculum Guide

Grade 5 Math Curriculum Guide Lesson 1 1 st Nine Weeks Unit 1: Number and Operations in Base Ten Understand Place Value 5.NBT.1. Recognize that in a multi-digit number, a digit in one place represents

### Section summaries. d = (x 2 x 1 ) 2 + (y 2 y 1 ) 2. 1 + y 2. x1 + x 2

Chapter 2 Graphs Section summaries Section 2.1 The Distance and Midpoint Formulas You need to know the distance formula d = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 and the midpoint formula ( x1 + x 2, y ) 1 + y 2

### Algorithm set of steps used to solve a mathematical computation. Area The number of square units that covers a shape or figure

Fifth Grade CCSS Math Vocabulary Word List *Terms with an asterisk are meant for teacher knowledge only students need to learn the concept but not necessarily the term. Addend Any number being added Algorithm

### CENTER OF GRAVITY, CENTER OF MASS AND CENTROID OF A BODY

CENTER OF GRAVITY, CENTER OF MASS AND CENTROID OF A BODY Dr. Amilcar Rincon-Charris, MSME Mechanical Engineering Department MECN 3005 - STATICS Objective : Students will: a) Understand the concepts of

### Solution Derivations for Capa #11

Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform

### Double integrals. Notice: this material must not be used as a substitute for attending the lectures

ouble integrals Notice: this material must not be used as a substitute for attending the lectures . What is a double integral? Recall that a single integral is something of the form b a f(x) A double integral

### RELATIVE MOTION ANALYSIS: VELOCITY

RELATIVE MOTION ANALYSIS: VELOCITY Today s Objectives: Students will be able to: 1. Describe the velocity of a rigid body in terms of translation and rotation components. 2. Perform a relative-motion velocity

### MODERN APPLICATIONS OF PYTHAGORAS S THEOREM

UNIT SIX MODERN APPLICATIONS OF PYTHAGORAS S THEOREM Coordinate Systems 124 Distance Formula 127 Midpoint Formula 131 SUMMARY 134 Exercises 135 UNIT SIX: 124 COORDINATE GEOMETRY Geometry, as presented

### Section 12.1 Translations and Rotations

Section 12.1 Translations and Rotations Any rigid motion that preserves length or distance is an isometry (meaning equal measure ). In this section, we will investigate two types of isometries: translations

### D Alembert s principle and applications

Chapter 1 D Alembert s principle and applications 1.1 D Alembert s principle The principle of virtual work states that the sum of the incremental virtual works done by all external forces F i acting in

### Interactive Math Glossary Terms and Definitions

Terms and Definitions Absolute Value the magnitude of a number, or the distance from 0 on a real number line Additive Property of Area the process of finding an the area of a shape by totaling the areas

### Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting

### PHYS 1111L LAB 2. The Force Table

In this laboratory we will investigate the vector nature of forces. Specifically, we need to answer this question: What happens when two or more forces are exerted on the same object? For instance, in

### Davis Buenger Math Solutions September 8, 2015

Davis Buenger Math 117 6.7 Solutions September 8, 15 1. Find the mass{ of the thin bar with density 1 x function ρ(x) 1 + x < x. Solution: As indicated by the box above, to find the mass of a linear object

### PLANE TRUSSES. Definitions

Definitions PLANE TRUSSES A truss is one of the major types of engineering structures which provides a practical and economical solution for many engineering constructions, especially in the design of

### Begin recognition in EYFS Age related expectation at Y1 (secure use of language)

For more information - http://www.mathsisfun.com/geometry Begin recognition in EYFS Age related expectation at Y1 (secure use of language) shape, flat, curved, straight, round, hollow, solid, vertexvertices

### Physics 1653 Exam 3 - Review Questions

Physics 1653 Exam 3 - Review Questions 3.0 Two uncharged conducting spheres, A and B, are suspended from insulating threads so that they touch each other. While a negatively charged rod is held near, but

### 9 Area, Perimeter and Volume

9 Area, Perimeter and Volume 9.1 2-D Shapes The following table gives the names of some 2-D shapes. In this section we will consider the properties of some of these shapes. Rectangle All angles are right

### Exam 2 Review. 3. How to tell if an equation is linear? An equation is linear if it can be written, through simplification, in the form.

Exam 2 Review Chapter 1 Section1 Do You Know: 1. What does it mean to solve an equation? To solve an equation is to find the solution set, that is, to find the set of all elements in the domain of the

### 39 Symmetry of Plane Figures

39 Symmetry of Plane Figures In this section, we are interested in the symmetric properties of plane figures. By a symmetry of a plane figure we mean a motion of the plane that moves the figure so that

### Handout 7: Magnetic force. Magnetic force on moving charge

1 Handout 7: Magnetic force Magnetic force on moving charge A particle of charge q moving at velocity v in magnetic field B experiences a magnetic force F = qv B. The direction of the magnetic force is

### Angles that are between parallel lines, but on opposite sides of a transversal.

GLOSSARY Appendix A Appendix A: Glossary Acute Angle An angle that measures less than 90. Acute Triangle Alternate Angles A triangle that has three acute angles. Angles that are between parallel lines,

### Physics 201 Homework 8

Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the

### MA.7.G.4.2 Predict the results of transformations and draw transformed figures with and without the coordinate plane.

MA.7.G.4.2 Predict the results of transformations and draw transformed figures with and without the coordinate plane. Symmetry When you can fold a figure in half, with both sides congruent, the fold line

### AP Physics - Chapter 8 Practice Test

AP Physics - Chapter 8 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A single conservative force F x = (6.0x 12) N (x is in m) acts on

### Newton s Third Law. object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1

Newton s Third Law! If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1!! Note on notation: is

### APPLICATION OF DERIVATIVES

6. Overview 6.. Rate of change of quantities For the function y f (x), d (f (x)) represents the rate of change of y with respect to x. dx Thus if s represents the distance and t the time, then ds represents

### AQA Level 2 Certificate FURTHER MATHEMATICS

AQA Qualifications AQA Level 2 Certificate FURTHER MATHEMATICS Level 2 (8360) Our specification is published on our website (www.aqa.org.uk). We will let centres know in writing about any changes to the

### Carroll County Public Schools Elementary Mathematics Instructional Guide (5 th Grade) August-September (12 days) Unit #1 : Geometry

Carroll County Public Schools Elementary Mathematics Instructional Guide (5 th Grade) Common Core and Research from the CCSS Progression Documents Geometry Students learn to analyze and relate categories

### Ground Rules. PC1221 Fundamentals of Physics I. Force. Zero Net Force. Lectures 9 and 10 The Laws of Motion. Dr Tay Seng Chuan

PC1221 Fundamentals of Physics I Lectures 9 and 10 he Laws of Motion Dr ay Seng Chuan 1 Ground Rules Switch off your handphone and pager Switch off your laptop computer and keep it No talking while lecture

### Example (1): Motion of a block on a frictionless incline plane

Firm knowledge of vector analysis and kinematics is essential to describe the dynamics of physical systems chosen for analysis through ewton s second law. Following problem solving strategy will allow

### E X P E R I M E N T 8

E X P E R I M E N T 8 Torque, Equilibrium & Center of Gravity Produced by the Physics Staff at Collin College Copyright Collin College Physics Department. All Rights Reserved. University Physics, Exp 8:

### Shape Dictionary YR to Y6

Shape Dictionary YR to Y6 Guidance Notes The terms in this dictionary are taken from the booklet Mathematical Vocabulary produced by the National Numeracy Strategy. Children need to understand and use

### Lesson 4 Rigid Body Statics. Taking into account finite size of rigid bodies

Lesson 4 Rigid Body Statics When performing static equilibrium calculations for objects, we always start by assuming the objects are rigid bodies. This assumption means that the object does not change

### acute angle acute triangle Cartesian coordinate system concave polygon congruent figures

acute angle acute triangle Cartesian coordinate system concave polygon congruent figures convex polygon coordinate grid coordinates dilatation equilateral triangle horizontal axis intersecting lines isosceles

### Version PREVIEW Practice 8 carroll (11108) 1

Version PREVIEW Practice 8 carroll 11108 1 This print-out should have 12 questions. Multiple-choice questions may continue on the net column or page find all choices before answering. Inertia of Solids

### Chapter 4 Dynamics: Newton s Laws of Motion

Chapter 4 Dynamics: Newton s Laws of Motion Units of Chapter 4 Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal

### Week #15 - Word Problems & Differential Equations Section 8.1

Week #15 - Word Problems & Differential Equations Section 8.1 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 25 by John Wiley & Sons, Inc. This material is used by

### Lecture 15. Torque. Center of Gravity. Rotational Equilibrium. Cutnell+Johnson:

Lecture 15 Torque Center of Gravity Rotational Equilibrium Cutnell+Johnson: 9.1-9.3 Last time we saw that describing circular motion and linear motion is very similar. For linear motion, we have position

### Physics-1 Recitation-3

Physics-1 Recitation-3 The Laws of Motion 1) The displacement of a 2 kg particle is given by x = At 3/2. In here, A is 6.0 m/s 3/2. Find the net force acting on the particle. (Note that the force is time

### Phys 102 Spg Exam No. 2 Solutions

Phys 102 Spg. 2008 Exam No. 2 Solutions I. (20 pts) A 10-turn wire loop measuring 8.0 cm by 16.0 cm carrying a current of 2.0 A lies in the horizontal plane and is free to rotate about a horizontal axis