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**Newton’s Laws of Motion**

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**Newton’s Laws of Motion**

The First, Second and Third Laws Mass, Force, Inertial Frame Weight, the Normal Force, Friction Centripetal Force and Circular Motion

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**Newton’s First Law of Motion**

This is Newton’s first law, which is often called the law of inertia: Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it. Figure 4-3. Caption: F represents the force applied by the person and Ffr represents the force of friction.

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**Newton’s First Law of Motion**

Inertial reference frames: Newton’s first law does not hold in every reference frame, such as a reference frame that is accelerating or rotating. An inertial reference frame is one in which Newton’s first law is valid. This excludes rotating and accelerating frames. How can we tell if we are in an inertial reference frame? By checking to see if Newton’s first law holds!

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Mass Mass is the measure of inertia of an object, sometimes understood as the quantity of matter in the object. In the SI system, mass is measured in kilograms. Mass is not weight. Mass is a property of an object. Weight is the force exerted on that object by gravity. If you go to the Moon, whose gravitational acceleration is about 1/6 g, you will weigh much less. Your mass, however, will be the same.

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**Newton’s Second Law of Motion**

Newton’s second law is the relation between acceleration and force. Acceleration is proportional to force and inversely proportional to mass. Figure 4-5. Caption: The bobsled accelerates because the team exerts a force.

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**Newton’s Third Law of Motion**

Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first. Figure 4-8. Caption: If your hand pushes against the edge of a desk (the force vector is shown in red), the desk pushes back against your hand (this force vector is shown in a different color, violet, to remind us that this force acts on a different object).

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**Newton’s Third Law of Motion**

Rocket propulsion can also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket. Note that the rocket does not need anything to “push” against. Figure Caption: Another example of Newton’s third law: the launch of a rocket. The rocket engine pushes the gases downward, and the gases exert an equal and opposite force upward on the rocket, accelerating it upward. (A rocket does not accelerate as a result of its propelling gases pushing against the ground.)

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A force is a push or pull. Force is a vector. Mass is the measure of inertia of an object. Mass is not weight. Mass is a property of an object. Weight is the force exerted on that object by gravity. An inertial reference frame is one in which Newton’s first law is valid. In the third law, the forces are exerted on different objects.

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**Weight and the Normal Force**

Weight is the force exerted on an object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight of an object of mass m is: where

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**Weight and the Normal Force**

The force exerted perpendicular to a surface is called the normal force. It is exactly as large as needed to balance the force from the object. Figure Caption: Example 4–6. (a) A 10-kg gift box is at rest on a table. (b) A person pushes down on the box with a force of 40.0 N. (c) A person pulls upward on the box with a force of 40.0 N. The forces are all assumed to act along a line; they are shown slightly displaced in order to be distinguishable. Only forces acting on the box are shown. Answer: The box is always at rest, so the forces must always add to zero. In (a), the normal force equals the weight; in (b), the normal force equals the weight plus the 40.0 N downward force; in (c), the normal force equals the weight minus the 40.0 N upward force.

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**Weight and the Normal Force**

Figure Caption: Example 4–6. (a) A 10-kg gift box is at rest on a table. (b) A person pushes down on the box with a force of 40.0 N. (c) A person pulls upward on the box with a force of 40.0 N. The forces are all assumed to act along a line; they are shown slightly displaced in order to be distinguishable. Only forces acting on the box are shown. Answer: The box is always at rest, so the forces must always add to zero. In (a), the normal force equals the weight; in (b), the normal force equals the weight plus the 40.0 N downward force; in (c), the normal force equals the weight minus the 40.0 N upward force.

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**Figure 4-19. Caption: Example 4–9:Two force vectors act on a boat.**

In this example, the net force on the boat has a magnitude of 53.3 N and acts at an 11° angle to the x axis.

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**A 65-kg woman descends in an elevator that briefly accelerates at 0**

A 65-kg woman descends in an elevator that briefly accelerates at 0.20g downward. She stands on a scale that reads in kg. During this acceleration, what is her weight and what does the scale read? (b) What does the scale read when the elevator descends at a constant speed of 2.0 m/s? Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors. Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg. The scale reads her true weight, or a mass of 65 kg.

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**Free-Body Diagrams Two boxes connected by a cord.**

Two boxes, A and B, are connected by a lightweight cord and are resting on a smooth table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force of 40.0 N is applied to the 10.0-kg box. Find (a) the acceleration of each box, and (b) the tension in the cord connecting the boxes. Figure Caption: Example 4–12. (a) Two boxes, A and B, are connected by a cord. A person pulls horizontally on box A with force FP = 40.0 N. (b) Free-body diagram for box A. (c) Free-body diagram for box B. Answer: Free-body diagrams for both boxes are shown. The net force on box A is the external force minus the tension; the net force on box B is the tension. Both boxes have the same acceleration. The acceleration is the external force divided by the total mass, or 1.82 m/s2. The tension in the cord is the mass of box B multiplied by the acceleration, or 21.8 N.

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**The advantage of a pulley. **

A mover is trying to lift a piano (slowly) up to a second-story apartment. He is using a rope looped over two pulleys as shown. What force must he exert on the rope to slowly lift the piano’s 2000-N weight? Figure 4-24. Answer: “Slowly” means with a constant speed, so the tension in the rope must be half the piano’s weight, or 1000 N.

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Accelerometer. A small mass m hangs from a thin string and can swing like a pendulum. You attach it above the window of your car as shown. What angle does the string make (a) when the car accelerates at a constant a = 1.20 m/s2, and (b) when the car moves at constant velocity, v = 90 km/h? Figure 4-25. Answer: (a) The acceleration of the mass is given by the horizontal component of the tension; the vertical component of the tension is equal to the weight. Writing these two equations and combining them gives tan θ = a/g, or θ = 7.0°. (b) When the velocity is constant, the string is vertical.

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**Box slides down an incline. **

A box of mass m is placed on a smooth incline that makes an angle θ with the horizontal. (a) Determine the normal force on the box. (b) Determine the box’s acceleration. (c) Evaluate for a mass m = 10 kg and an incline of θ = 30°. Figure Caption: Example 4–16. (a) Box sliding on inclined plane. (b) Free-body diagram of box. Answer: On an incline (or any surface), the normal force is perpendicular to the surface and any frictional forces are parallel to the surface. (a) The normal force is equal to the component of the weight perpendicular to the incline, or mg cos θ. (b) The force causing the acceleration is the component of the weight parallel to the incline; therefore the acceleration is g sin θ. (c) The normal force is 85 N and the acceleration is 4.9 m/s2.

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**Newton’s Laws Involving Friction**

Approximation of the frictional force: Ffr = μkFN . Here, FN is the normal force, and μk is the coefficient of friction, which is different for each pair of surfaces.

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**Friction: static and kinetic**

Our 10.0-kg mystery box rests on a horizontal floor. The coefficient of static friction is 0.40 and the coefficient of kinetic friction is Determine the force of friction acting on the box if a horizontal external applied force is exerted on it of magnitude: (a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N. Figure 5-2. Figure 5-3. Caption: Example 5–1. Magnitude of the force of friction as a function of the external force applied to an object initially at rest. As the applied force is increased in magnitude, the force of static friction increases linearly to just match it, until the applied force equals μsFN. If the applied force increases further, the object will begin to move, and the friction force drops to a roughly constant value characteristic of kinetic friction. Solution: Since there is no vertical motion, the normal force equals the weight, 98.0 N. This gives a maximum static frictional force of 29 N. Therefore, for parts a-d, the frictional force equals the applied force. In (e), the frictional force is 29 N and the box accelerates at 1.1 m/s2 to the right.

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**A 10. 0-kg box is pulled along a horizontal surface by a force of 40**

A 10.0-kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is Calculate the acceleration. Figure 5-5. Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.

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**Newton’s Laws Involving Friction**

Will you exert less force if you push her or pull her? Assume the same angle θ in each case. Figure 5-6. Answer: Pulling decreases the normal force, while pushing increases it. Better pull.

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**Newton’s Laws Involving Friction**

Two boxes and a pulley. Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is We ignore the mass of the cord and pulley and any friction 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 B moves down, box A moves to the right. Figure 5-7. Solution: For box A, the normal force equals the weight; we therefore know the magnitude of the frictional force, but not of the tension in the cord. Box B has no horizontal forces; the vertical forces on it are its weight (which we know) and the tension in the cord (which we don’t). Solving the two free-body equations for a gives 1.4 m/s2 (and the tension as 17 N).

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The skier. This skier is descending a 30° slope, at constant speed. What can you say about the coefficient of kinetic friction? Figure 5-8. Caption: Example 5–6. A skier descending a slope; FG = mg is the force of gravity (weight) on the skier. Solution: Since the speed is constant, there is no net force in any direction. This allows us to find the normal force and then the frictional force; the coefficient of kinetic friction is 0.58.

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**Newton’s Laws Involving Friction**

Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. (a) If the coefficient of static friction is 0.40, determine what range of values for mass B will keep the system at rest. (b) If the coefficient of kinetic friction is 0.30, and mB = 10.0 kg, determine the acceleration of the system. Figure 5-9. Caption: Example 5–7. Note choice of x and y axes. Solution: Pick axes along and perpendicular to the slope. Box A has no movement perpendicular to the slope; box B has no horizontal movement. The accelerations of both boxes, and the tension in the cord, are the same. Solve the resulting two equations for these two unknowns. The mass of B has to be between 2.8 and 9.2 kg (so A doesn’t slide either up or down). 0.78 m/s2

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**Uniform Circular Motion**

Motion in a circle of constant radius at constant speed. Instantaneous velocity is always tangent to the circle. This acceleration is called the centripetal, or radial, acceleration, and it points toward the center of the circle. Figure Caption: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

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**Uniform Circular Motion**

Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that Figure Caption: Determining the change in velocity, Δv, for a particle moving in a circle. The length Δl is the distance along the arc, from A to B.

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**Dynamics of Uniform Circular Motion**

For an object to be in uniform circular motion, there must be a net force acting on it. We already know the acceleration, so can immediately write the force: Figure Caption: A force is required to keep an object moving in a circle. If the speed is constant, the force is directed toward the circle’s center.

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**Dynamics of Uniform Circular Motion**

We can see that the force must be inward by thinking about a ball on a string. Strings only pull; they never push. Figure Caption: Swinging a ball on the end of a string.

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**Dynamics of Uniform Circular Motion**

There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome. If the centripetal force vanishes, the object flies off at a tangent to the circle. Figure Caption: If centrifugal force existed, the revolving ball would fly outward as in (a) when released. In fact, it flies off tangentially as in (b). For example, in (c) sparks fly in straight lines tangentially from the edge of a rotating grinding wheel.

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**Dynamics of Uniform Circular Motion**

Force on revolving ball (horizontal). Estimate the force a person must exert on a string attached to a kg ball to make the ball revolve in a horizontal circle of radius m. The ball makes 2.00 revolutions per second. Ignore the string’s mass. Figure 5-17. Answer: Ignoring the weight of the ball, so FT is essentially horizontal, we find FT = 14 N.

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**Dynamics of Uniform Circular Motion**

Revolving ball (vertical circle). A kg ball on the end of a 1.10-m-long cord (negligible mass) is swung in a vertical circle. (a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. (b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the speed of part (a). Figure Caption: Example 5–12. Freebody diagrams for positions 1 and 2. Solution: See the freebody diagrams. The minimum speed occurs when the tension is zero at the top; v = m/s. Substitute to find FT2 = 7.35 N.

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**Dynamics of Uniform Circular Motion**

Conical pendulum. A small ball of mass m, suspended by a cord of length l, revolves in a circle of radius r = l sin θ, where θ is the angle the string makes with the vertical. (a) In what direction is the acceleration of the ball, and what causes the acceleration? (b) Calculate the speed and period (time required for one revolution) of the ball in terms of l, θ, g, and m. Figure Caption: Example 5–13. Conical pendulum. Answer: a. The acceleration is towards the center of the circle, and it comes from the horizontal component of the tension in the cord. b. See text.

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Skidding on a curve. Figure Caption: Example 5–14. Forces on a car rounding a curve on a flat road. (a) Front view, (b) top view. Solution: The normal force equals the weight, and the centripetal force is provided by the frictional force (if sufficient). The required centripetal force is 4500 N. The maximum frictional force is 5880 N, so the car follows the curve. The maximum frictional force is 2450 N, so the car will skid.

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**Banking the curve can help keep cars from skidding**

Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed at which the entire centripetal force is supplied by the horizontal component of the normal force, and no friction is required. This occurs when: Figure Caption: Normal force on a car rounding a banked curve, resolved into its horizontal and vertical components. The centripetal acceleration is horizontal (not parallel to the sloping road). The friction force on the tires, not shown, could point up or down along the slope, depending on the car’s speed. The friction force will be zero for one particular speed.

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