今日課程內容CH4: 牛頓運動定律

牛頓第二運動定律牛頓第三運動定律物理上常見的力

CH5: 牛頓運動定律 ( 應用 )摩擦力等速率圓周運動

4-4 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.

It takes a force to change either the direction or the speed of an object. More force means more acceleration; the same force exerted on a more massive object will yield less acceleration.

4-4 Newton’s Second Law of Motion

Force is a vector, so is true along each coordinate axis.

The unit of force in the SI system is the newton (N).

Note that the pound is a unit of force, not of mass, and can therefore be equated to newtons but not to kilograms.

4-4 Newton’s Second Law of Motion

Example 4-2: Force to accelerate a fast car.

Estimate the net force needed to accelerate (a) a 1000-kg car at ½ g; (b) a 200-g apple at the same rate.

Example 4-3: Force to stop a car.

What average 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?

4-5 Newton’s Third Law of Motion

Any time a force is exerted on an object, that force is caused by another object.

Newton’s third law:

Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.

4-5 Newton’s Third Law of Motion

A key to the correct application of the third law is that the forces are exerted on different objects. Make sure you don’t use them as if they were acting on the same object.

When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction.

4.5: Newton’s Third Law

• The minus sign means that these two forces are in opposite directions

• The forces between two interacting bodies are called a third-law force pair. • 內力與外力

4-5 Newton’s Third Law of Motion

Helpful notation: the first subscript is the object that the force is being exerted on; the second is the source.

4-6 Weight—the Force of Gravity; 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

4-6 Weight—the Force of Gravity; and the Normal Force( 正向力 )

An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there?

The force exerted perpendicular to a surface iscalled the normal force. It is exactly as large as needed to balance the force from the object. (If the required force gets too big, something breaks!)

Tension( 張力 )When a cord is attached to a body and pulled taut, the cord

pulls on the body with a force T directed away from the body and along the cord.

Fig. 5-9 (a) The cord, pulled taut, is under tension. If its mass is negligible, the cord pulls on the body and the hand with force T, even if the cord runs around a massless, frictionless pulley as in (b) and (c).

4-6 Weight—the Force of Gravity; and the Normal Force

Example 4-6: Weight, normal force, and a box.

A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table.

(a) Determine the weight of the box and the normal force exerted on it by the table.

(b) Now your friend pushes down on the box with a force of 40.0 N. Again determine the normal force exerted on the box by the table.

(c) If your friend pulls upward on the box with a force of 40.0 N, what now is the normal force exerted on the box by the table?

4-6 Weight—the Force of Gravity; and the Normal Force

Example 4-7: Accelerating the box.

What happens when a person pulls upward on the box in the previous example with a force greater than the box’s weight, say 100.0 N?

4-6 Weight—the Force of Gravity; and the Normal Force

Example 4-8: Apparent weight loss.

A 65-kg woman descends in an elevator that briefly accelerates at 0.20g downward. She stands on a scale that reads in kg.

(a) 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?

4-7 Solving Problems with Newton’s Laws: Free-Body Diagrams

Example 4-12: 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.

4-7 Solving Problems with Newton’s Laws: Free-Body Diagrams

Example 4-15: 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?

4-7 Solving Problems with Newton’s Laws: Free-Body Diagrams

Example 4-16: 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°.

5-1 Applications of Newton’s Laws Involving Friction( 磨擦力 )

Friction is always present when two solid surfaces slide along each other.

The microscopic details are not yet fully understood.

5-1 Applications of Newton’s Laws Involving Friction

Sliding friction is called kinetic friction( 動磨擦力 ).

Approximation of the frictional force:

Ffr = μkFN .

Here, FN is the normal force, and μk is the coefficient of kinetic friction, which is different for each pair of surfaces.

5-1 Applications of Newton’s Laws Involving Friction

Static friction( 靜磨擦力 ) applies when two surfaces are at rest with respect to each other (such as a book sitting on a table).

The static frictional force is as big as it needs to be to prevent slipping, up to a maximum value.

Ffr ≤ μsFN .

Usually it is easier to keep an object sliding than it is to get it started.

5-1 Applications of Newton’s Laws Involving Friction

Example 5-1: 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 0.30. 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.

5-1 Applications of Newton’s Laws Involving Friction

Conceptual Example 5-2: A box against a wall.

You can hold a box against a rough wall and prevent it from slipping down by pressing hard horizontally. How does the application of a horizontal force keep an object from moving vertically?

5-1 Applications of Newton’s Laws Involving Friction

Example 5-3: Pulling against friction.

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 0.30. Calculate the acceleration.

5-1 Applications of Newton’s Laws Involving Friction

Example 5-7: A ramp, a pulley, and two boxes.

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.

5-2 Uniform Circular Motion—KinematicsUniform circular motion: motion in a circle of constant radius at constant speed

Instantaneous velocity is always tangent to the circle.

2 Rv R

t T

d

dt

1801rad弧度 (rad)

Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that

5-2 Uniform Circular Motion—Kinematics

.

This acceleration is called the centripetal, or radial, acceleration( 向心或徑向加速度 ), and it points toward the center of the circle.

5-2 Uniform Circular Motion—Kinematics

2 22

2

4C

v Ra R

R T

R

5-2 Uniform Circular Motion—Kinematics

Example 5-8: Acceleration of a revolving ball.

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

5-3 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:

5-3 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.

5-3 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.

5-3 Dynamics of Uniform Circular Motion

Example 5-11: Force on revolving ball (horizontal).

Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.

5-3 Dynamics of Uniform Circular Motion

Example 5-12: Revolving ball (vertical circle).

A 0.150-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).

5-3 Dynamics of Uniform Circular Motion

Example 5-13: 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.

5-4 Highway Curves: Banked and Unbanked

When a car goes around a curve, there must be a net force toward the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

r

5-4 Highway Curves: Banked and Unbanked

Example 5-14: Skidding on a curve.

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s (54 km/h). Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is μs = 0.60; (b) the pavement is icy and μs = 0.25.

5-4 Highway Curves: Banked and Unbanked

If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.

5-4 Highway Curves: Banked and Unbanked

As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways:

1. The kinetic frictional force is smaller than the static.

2. The static frictional force can point toward the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.

5-4 Highway Curves: Banked and Unbanked

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:

習題Ch1: 4, 7, 8, 17, 22, Ch2: 3, 6, 15, 18, 24, 27, 37, 43, 60, 67,Ch3: 8, 9, 24, 25, 38, 44, 48, 60, 61, 67