Chapter 4A. Translational Equilibrium. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University
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1 Chapter 4. Translational Equilibrium PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007
2 MOUNTIN CLIMER exerts action forces on crevices and ledges, which produce reaction forces on the climber, allowing him to scale the cliffs. Photo by Photo Disk Vol. 1/Getty
3 Objectives: fter completing this module, you should be able to: State and describe with examples Newton s three laws of motion. State and describe with examples your understanding of the first condition for equilibrium. Draw free-body diagrams for objects in translational equilibrium. Write and apply the first condition for equilibrium to the solution of problems similar to those in this module.
4 Newton s s First Law Newton s s First Law: n object at at rest or or an object in in motion at at constant speed will remain at at rest or or at at constant speed in in the absence of of a resultant force. glass is is placed on a board and the board is is jerked quickly to the right. The glass tends to remain at rest while the board is is removed.
5 Newton s s First Law (Cont.) Newton s s First Law: n object at at rest or or an object in in motion at at constant speed will remain at at rest or or at at constant speed in in the absence of of a resultant force. ssume glass and board move together at constant speed. If If the board stops suddenly, the glass tends to maintain its constant speed.
6 Understanding the First Law: Discuss what the driver experiences when a car accelerates from rest and then applies the brakes. (a) The driver is forced to move forward. n object at rest tends to remain at rest. (b) Driver must resist the forward motion as brakes are applied. moving object tends to remain in motion.
7 Newton s s Second Law Newton s second law of motion will be discussed quantitatively in in a later chapter, after we have covered acceleration. cceleration is is the rate at which the speed of an object changes. n object with an acceleration of 2 m/s 2, for example, is is an object whose speed increases by 2 m/s every second it it travels.
8 Newton s s Second Law: Second Law: Whenever a resultant force acts on an object, it it produces an acceleration - an acceleration that is is directly proportional to the force and inversely proportional to the mass. a F m
9 cceleration and Force With Zero Friction Forces Pushing the cart with twice the force produces twice the acceleration. Three times the force triples the acceleration.
10 cceleration and Mass gain With Zero Friction F F a a/2 Pushing two carts with same force F produces one-half the acceleration. The acceleration varies inversely with the amount of material (the mass).
11 Newton s s Third Law To every action force there must be an equal and opposite reaction force. Force of Ceiling on Man Force of Man on Ceiling Force of Floor on Man Force of Man on Floor Force of Wall on Hands Force of Hands on Wall ction and reaction forces act on different objects.
12 Newton s s Third Law Two More Examples: ction Reaction Reaction ction ction and Reaction Forces ct on Different Objects. They Do Not Cancel Each Other!
13 Translational Equilibrium n object is said to be in Translational Equilibrium if and only if there is no resultant force. This means that the sum of all acting forces is zero. C In the example, the resultant of of the three forces,,,, and C acting on the ring must be zero.
14 Visualization of Forces Force diagrams are necessary for studying objects in equilibrium. Don t confuse action forces with reaction forces. Equilibrium: F 0 The action forces are each ON the ring. C Force : y ceiling on ring. Force : y ceiling on ring. Force C: y weight on ring.
15 Visualization of Forces (Cont.) Now let s look at the Reaction Forces for the same arrangement. They will be equal, but opposite, and they act on different objects. Reaction forces: Reaction forces are each exerted: Y the ring. r C r r Force r : y ring on ceiling. Force r : y ring on ceiling. Force C r : y ring on weight.
16 Vector Sum of Forces n object is said to be in Translational Equilibrium if and only if there is no resultant force. The vector sum of all forces acting on the ring is zero in this case. W C 40 0 Vector sum: F = + + C = 0
17 Vector Force Diagram W C 40 0 y 40 0 y W C x free-body diagram is a force diagram showing all the elements in this diagram: axes, vectors, components, and angles.
18 Free-body Diagrams: Read problem; draw and label sketch. Isolate a common point where all forces are acting. Construct force diagram at at origin of of x,y axes. Dot in in rectangles and label x and y components opposite and adjacent to to angles. Label all given information and state what forces or or angles are to to be found.
19 Look gain at Previous rrangement 40 0 y 40 0 y W C W C x 1. Isolate point. 2. Draw x,y axes. 3. Draw vectors. 4. Label components. 5. Show all given information.
20 Example 1. Draw a free-body diagram for the arrangement shown on left. The pole is light and of negligible weight. On rope W 30 0 C 700 N Careful: The pole can only push or pull since it has no weight. y 30 0 C 700 N x The force is is the force exerted on the rope Isolate the rope at the end of the boom. by the pole. Don t confuse it it with the reaction ll forces must act ON the rope! force exerted by the rope on the pole.
21 Translational Equilibrium The First Condition for Equilibrium is that there be no resultant force. This means that the sum of all acting forces is zero. F 0 F 0 x y
22 Example 2. Find the tensions in ropes and for the arrangement shown. C 40 0 y 40 0 y 200 N C x 200 N The Resultant Force on the ring is zero: R = F = 0 R x = x + x + C x = 0 R y = y + y + C y = 0
23 Example 2. (Cont.) Finding components. y = 0 Recall trigonometry to find components: y 40 0 x x C C C = 0 y x dj = Hyp x cos x = cos 40 0 Opp= Hypx sin y = sin 400 The components of the vectors are found from the freebody diagram. 200 N C y = -200 N
24 Example 2. Continued... Components x = cos 40 0 y = sin 40 0 x = ; y = 0 C x = 0; C y = W y W C 40 0 x y free-body diagram must represent all forces as components along x and y-axes. It It must also show all given information.
25 Example 2. Continued... C N y 200 N 40 0 y C x F x = 0 F y = 0 Components x = cos 40 0 y = sin 40 0 x = ; y = 0 C x = 0; C y = W 0 0 F cos40 0; or cos40 x Fy 0 0 sin N 0; or sin N
26 Example 2. Continued... y 200 N 40 0 y C x Two equations; two unknowns sin N cos 40 0 Solve first 200 N N for sin 40 Solve Next for 0 0 cos40 (311 N)cos40 ; =238 N The tensions in and are = 311 N; = 238 N
27 Problem Solving Strategy 1. Draw a sketch and label all information. 2. Draw a free-body diagram. 3. Find components of all forces (+ and -). 4. pply First Condition for Equilibrium: F x = 0 ; F y = 0 5. Solve for unknown forces or angles.
28 Example 3. Find Tension in Ropes and y y 400 N x 400 N x 1. Draw free-body diagram. 2. Determine angles. 3. Draw/label components. Next we will find components of each vector.
29 Example 3. Find the tension in ropes and. First Condition for Equilibrium: y y F x = 0 ; F y = 0 x x W 400 N 4. pply 1 st Condition for Equilibrium: F x = x - x = 0 F y = y + y -W = 0 x = x y + y = W
30 Example 3. Find the tension in ropes and. x = cos 30 0 ; y = sin 30 0 x = cos 60 0 y = sin 60 0 y W x = 0; W y = -400 N W 400 N x x y Using Trigonometry, the first condition yields: x = x y + y = W cos 60 0 = cos 30 0 sin sin 60 0 = 400 N
31 Example 3 (Cont.) Find the tension in and. y x x y cos 60 0 = cos 30 0 sin sin 60 0 = 400 N W 400 N We now solve for and : Two Equations and Two Unknowns. We will first solve the horizontal equation for in terms of the unknown : 0 cos = cos 60
32 Example 3 (Cont.) Find Tensions in and. y x x 400 N y = Now apply Trig to: y + y = 400 N sin sin 60 0 = 400 N = sin sin 60 0 = 400 N sin (1.732 ) sin 60 0 = 400 N = 400 N = 200 N
33 Example 3 (Cont.) Find with = 200 N. y x x W 400 N y = 200 N = = 1.732(400 N) = 346 N Rope tensions are: = 200 N and = 346 N This problem is made much simpler if you notice that the angle between vectors and is 90 0 and rotate the x and y axes (Continued)
34 Example 4. Rotate axes for same example y y x y 400 N x 400 N x W We recognize that and are at right angles, and choose the x-axis along not horizontally. The y-axis will then be along with W offset.
35 Since and are perpendicular, we can find the new angle from geometry. y x y x N W =400 N You should show that the angle will be We now only work with components of W.
36 Recall W = 400 N. Then we have: y W x 30 0 x W x = (400 N) cos 30 0 W y = (400 N) sin 30 0 Thus, the components of the weight vector are: W y 400 N W x = 346 N; W y = 200 N pply the first condition for Equilibrium, and... W x x = 0 and W y y = 0
37 Example 4 (Cont.) We Now Solve for and : y x F x = - W x = 0 = W x = (400 N) cos 30 0 W x 30 0 W y 400 N efore working a problem, you might see if rotation of the axes helps. = 346 N F y = - W y = 0 = W y = (400 N) sin 30 0 = 200 N
38 Summary Newton s s First Law: n object at rest or an object in in motion at constant speed will remain at rest or at constant speed in in the absence of a resultant force.
39 Summary Second Law: Whenever a resultant force acts on an object, it it produces an acceleration, an acceleration that is is directly proportional to the force and inversely proportional to the mass.
40 Summary Third Law: To every action force there must be an equal and opposite reaction force. ction Reaction Reaction ction
41 Free-body Diagrams: Read problem; draw and label sketch. Isolate a common point where all forces are acting. Construct force diagram; t origin of of x,y axes. Dot in in rectangles and label x and y components opposite and adjacent to to angles. Label all given information and state what forces or or angles are to to be found.
42 Translational Equilibrium The First Condition for Equilibrium is that there be no resultant force. This means that the sum of all acting forces is zero. F 0 F 0 x y
43 Problem Solving Strategy 1. Draw a sketch and label all information. 2. Draw a free-body diagram. 3. Find components of all forces (+ and -). 4. pply First Condition for Equilibrium: F x = 0 ; F y = 0 5. Solve for unknown forces or angles.
44 Conclusion: Chapter 4 Translational Equilibrium
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