MOTION AND FORCE: DYNAMICS

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1 MOTION AND FORCE: DYNAMICS We ve been dealing with the fact that objects move. Velocity, acceleration, projectile motion, etc. WHY do they move? Forces act upon them, that s why! The connection between Force and motion is DYNAMICS. FORCE A push or pull on an object. An object falls due to the force of gravity. Forces do not always give rise to motion.! Spring Scale--a device used to measure force! Weight--a measure of the FORCE on an object, carries a unit of Newtons (N).! Forces--are vectors that carry a magnitude and direction. NEWTON S FIRST LAW OF MOTION [Law of Inertia] An object in motion remains in motion along a straight line path until a force acts upon that object. An object at rest remains at rest until a force acts upon that object.! Inertia--tendency of a body to maintain its state of rest or uniform motion.! Friction--a force that acts against motion! Inertial Reference Frame--Newton s first law holds true. Rest a cup on a dashboard in your car before leaving your garage--back out of the garage-- splash! That is NOT an inertial frame of reference!! Mass--a property; measure of the inertia of a body (kg)! Weight--a force; F g = mg measured in Newtons. A word that has a casual meaning on Earth since g is relatively 9.8 m/s 2. Mass and weight have often been used interchangeably during your short lifetime! NEWTON S SECOND LAW OF MOTION [F = ma] The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object. a = EF m In English: When a net force acts upon an object, it s rest or uniform motion is changed. That means it experiences an acceleration, technically defined as a change in velocity.! E F = ma! E F = net force! Force--an action capable of accelerating an object; measured in Newtons! 1 N = kg Cm/s 2! ONE Dimensional L F = ma! TWO Dimensional L F x = ma x & F y = ma y & F z = ma z

2 Example 4.1 Estimate the net force needed to accelerate a 1000-kg car at 1/2 g. Example 4.2 What net force is required to bring a 1500-kg car to rest from a speed of 100 km/h within a distance of 55 m? NEWTON S THIRD LAW [action-reaction] Whenever on object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. Often stated, To every action there is an equal and opposite reaction. The second law quantifies how forces affect motion. But, where do forces come from? A force is exerted ON an object and BY another object. Hit a nail with a hammer--the force is applied ON the nail BY the hammer BUT, the nail exerts a force back on the hammer. How do we know? 1) the speed of the hammer is rapidly reduced to zero! 2) this force must be strong to so rapidly reduce the speed of the hammer 3) the nail rarely sinks all the way into the wood on the first try! Example 4.3 What makes a car go forward? Forces 2

3 Example 4.4 Michelangelo s assistant has been assigned the task of moving a block of marble using a sled. He says to his boss, When I exert a forward force on the sled, the sled exerts an equal and opposite force backward, so how can I ever start it moving? No matter how hard I pull, the backward reaction force always equals my forward force, so the net force must be zero. I ll never be able to move this load. Is this a case of a little knowledge being dangerous? Explain. WEIGHT--THE FORCE OF GRAVITY; AND THE NORMAL FORCE! Weight--The force of gravity originates at the center of the Earth. F g = F w = mg always directed downward, toward the center of the Earth.! g--9.8 m/s 2 The moon s gravity is 1/6 that of the Earth since the moon s mass is 1/6 that of the Earth s. Your weight on the moon is 1/6 that of your weight here on Earth--I find this a pleasant thought! So why don t we go crashing through the floor on to the poor souls below? The force of the floor exerts force right back at ya due to it s elasticity. This force is the normal force and is NOT a case of action-reaction!! Normal force-- normal means z (usually thought of as z to the motion!). F N! Since E F = 0 for an object at rest and g is acting downward this force often balances the force of weight for a stationary object. Forces 3

4 Example 4.5 A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. It s a reward for your fine showing on the physics final. The box is resting on a smooth (frictionless) horizontal surface of a table. a) Determine the weight of the box and the normal force acting on it. b) Now your friend pushes down on the box with a force of 40.0 N. Again determine the normal force acting on the box. c) If your friend pulls upward on the box with a force of 40.0 N, what now is the normal force on the box? Notice that the normal force is elastic in origin! THAT S WHY IS NOT AN ACTION-REACTION PAIR WITH THE WEIGHT! Example 4.6 What happens when a person pulls upward on the box in Example 4.5 c) with a force equal to, or greater than, the box s weight, say F p = N rather than the 40.0 N shown in figure 4-16 c)? Forces 4

5 SOLVING PROBLEMS WITH NEWTON S LAWS: VECTOR FORCES AND FREE-BODY DIAGRAMS Net force = vector sum = E F and is % to acceleration of an object Example 4.7 Calculate the sum of the two forces acting on the boat shown in Fig. 4-19a. Free-Body diagram:! Show all forces acting on EACH object involved! Point masses for now... until we start rotating stuff! Example 4.8 A hockey puck is sliding at constant velocity across a flat horizontal ice surface that is assumed to be frictionless. Which of the sketches is the correct free-body diagram for this puck? What would your answer be if the puck was slowed down? Forces 5

6 Example 4.9 Suppose a friend asks to examine the 10.0 kg box you were given earlier, hoping to guess what s inside. You respond, Sure, pull the box over to you. She then pulls the box over by an attached ribbon/string/cord/rope along the smooth surface of the table. The magnitude of the force exerted by the person is 40.0 N and is exerted at a 30.0 angle with the table top. Calculate a) the acceleration of the box. b) the magnitude of the upward force, F N exerted by the table on the box. Assume that friction can be neglected.! Tension--F T --when a flexible cord/rope/etc. pulls on an object it is said to be under tension. These can only pull, NOT push since they are flexible pieces of matter! Example 4.10 Two boxes are connected by a lightweight cord and are resting on a table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force F p of 40.0 N is applied by a person to the 10.0 kg box. Find a) the acceleration of each box Forces 6

7 b) the tension in the cord. Example 4.11 Two masses suspended over a pulley by a cable is sometimes referred to as an Atwood s machine. Consider the real-life application of an elevator (m 1 ) and its counterweight (m 2 ). To minimize the work done by the motor to raise and lower the elevator safely, m 1 and m 2 are similar in mass. We leave the motor out of the system for this calculation, and assume the cable s mass is negligible and the pulley is frictionless and massless, which assures that the tension, F T, in the cord has the same magnitude on both sides of the pulley. Let the mass of the counterweight be m 2 = 1,000 kg. Assume the mass of the empty elevator is 850 kg, and its mass when carrying 4 passengers is m 1 = 1150 kg. For the latter case (m 1 = 1150 kg), calculate a) the acceleration of the elevator and b) the tension in the cable. Forces 7

8 Example 4.12 Muscleman is trying to lift a piano (slowly) up to a second-story apartment. He is using a rope looped over two pulleys as shown. How much of the piano s 2000 N weight does he have to pull on the rope? Example 4.13 Finding her car stuck in the mud, a bright graduate student of a good physics course ties a strong rope to the back bumper of the car, and the other end to a tree. She pushes at the midpoint of the rope with her maximum effort, which she estimates to be a force F p. 300 N. The car just begins to budge with the rope at an angle 2 which she estimates to be 5. With what force is the rope pulling on the car? Neglect the mass of the rope. Forces 8

9 APPLICATIONS INVOLVING FRICTION; INCLINES Friction--exists between 2 solid surfaces because even the smoothest looking surface is quite rough on a microscopic scale. When we try to slide an object across another surface, these microscopic bumps impede motion. Additionally, the atoms snuggle up next to one another and have interactions of an attractive nature and can further impede motion. kinetic friction--sliding friction; the friction that persists even once the object is in motion. This force acts opposite to the body s velocity and is determined by the nature of the two surfaces. The force of kinetic friction is approximately proportional to the normal force. [the normal force is the force that either object exerts on the other perpendicular to their common surface of contact] F f % F N insert a proportionality constant and presto, an equals sign appears along with a constant! F f = : K F N : K --the coefficient of kinetic friction; its value depends on the nature of the 2 surfaces; it s an approximation since polishing/ sanding, etc. alters the surfaces. This is not a vector equation since the two vectors act z to one another. It is also an experimental relationship NOT a fundamental law. The force of friction depends very little on surface area. It doesn t matter whether you slide your book flat along the table or on its spine along the table--it s frictional force is essentially unaffected. Forces 9

10 Static friction--force applied to get an object moving; it is always present between 2 stationary objects and increases when the force applied increases until the force applied overpowers the static frictional force and the kinetic frictional force takes over. At the point it begins to move you have applied the maximum force of static friction F MAX # : S F N than get it to move! you ve probably noticed it s easier to keep an object moving Example 4.14 Our 10.0 kg mystery box rests on a horizontal floor. The coefficient of static friction is : s = 0.40 and the coefficient of kinetic friction is : K = Determine the force of friction, F f, acting on the box if a horizontal external applied force F A is exerted on it of magnitue [answer in 2 sig. figs for each part!] a) 0 N b) 10 N c) 20 N d) 38 N e) 40 N Forces 10

11 Notice that BOTH the normal force and the frictional forces are exerted by one surface ON another. F N is perpendicular to the contact surface F f is parallel to the contact surface and in the opposite direction of the x velocity. Example 4.15 Your little sister wants a ride on her sled. If you are on flat ground, will ou exert less force if you push her or pull her? Assume the same angle 2 in each case. Example 4.16 Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box I and the table is We ignore the mass of the cord and pulley and any fricion in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn t stretch. As box II moves down, box I moves to the right. Forces 11

12 OBJECTS MOVING ON AN INCLINE Easier to choose the xy coordinate system so that the x-axis is parallel to the incline surface and the y-axis is therefore, perpendicular to the incline. This helps because then a has only one component and if friction is present, two of the forces will have only one component: F f along the plane, opposite ot the object s velocity, and F N which is NOT vertical but is perpendicular to the plane. A picture will help! FN = FW cosθ F = F sinθ = mgsinθ applied W Example 4.17 A skier has just begun descending a 30.0 slope. Assuming the coefficient of kinetic friction is 1.10, a) draw the free body diagram b) calculate her acceleration c) calculate the speed she will reach after 4.0 seconds. Example 4.18 Suppose the snow is slushy and the skier moves down the 30.0 slope at constant speed. Determine the coefficient of friction, : K? Forces 12

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