Lecture 7. Forces. Important note: First Exam is next Tuesday, Feb. 6, 8:15-9:45 pm (see link on Canvas for locations)

Transcription

1 Lecture 7 Forces Important note: First Exam is next Tuesday, Feb. 6, 8:15-9:45 pm (see link on Canvas for locations)

2 Today s Topics: Forces The gravitational force The normal force Frictional Forces Next lecture. Lots of examples

3 Newton s Law of Gravitation Every particle exerts an attractive force on every other particle. A particle is a piece of matter, small enough in size to be regarded as a mathematical point. The force that each exerts on the other is directed along the line joining the particles. F = G m m 1 r 2 2 G = N m 2 kg 2

4 Defining weight The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downwards, toward the center of the earth. On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. W = G M E r m 2 W = mg g = G M r E 2

5 At the earth s surface: g = = G M R E 2 E ( ) ( 24 ) kg N m kg ( 6 ) m = 9.80 m s 2

6 ACT: Stone and feather I What can you say about the force of gravity F g acting on a stone and a feather on the moon? a) F g is greater on the feather b) F g is greater on the stone c) F g is zero on both due to vacuum d) F g is equal on both always e) F g is zero on both always The force of gravity (weight) depends on the mass of the object!! The stone has more mass, and therefore more weight.

7 ACT: Stone and feather II What can you say about the acceleration of gravity acting on the stone and the feather? a) it is greater on the feather b) it is greater on the stone c) it is zero on both due to vacuum d) it is equal on both always e) it is zero on both always The acceleration is given by F/m so here the mass divides out. Because we know that the force of gravity (weight) is mg, then we end up with acceleration g for both objects. Follow-up: Which one hits the bottom first?

8 Defining the Normal force The normal force is one component of the force that a surface exerts on an object with which it is in contact namely, the component that is perpendicular to the surface.

9 Block on two tables A block is balanced in the space between two tables as shown below. What forces are acting on the block? N by left table N by right table! F net = 0 W by Earth N R = N L = W 2 (½ with enough symmetry) If the block rests on 100 mini-tables, each table exerts a relatively small force: DEMO: Nail bed N each table = W 100

10 ACT: Contact force If you push with force F on either the heavy box (m 1 ) or the light box (m 2 ), in which of the two cases below is the contact force between the two boxes larger? a) case A b) case B c) same in both cases The acceleration of both masses together is the same in either case. But the contact force is the only force that accelerates m 1 in case A (or m 2 in case B). Because m 1 is the larger mass, it requires the larger contact force to achieve the same acceleration. A F B m m 1 2 m 2 m 1 F

11 ACT: Normal force Below you see two cases: a man pulling or pushing a sled with a force F that is applied at an angle θ. In which case is the normal force greater? a) case 1 b) case 2 c) it s the same for both d) depends on the magnitude of the force F 5) depends on the ice surface In case 1, the force F is pushing down (in addition to mg), so the normal force Case 1 needs to be larger. In case 2, the force F is pulling up, against gravity, so the Case 2 normal force is lessened.

12 Apparent Weight å Fy = + FN - mg = ma F = mg + N ma apparent weight true weight

13 Inclined plane (no friction) F N F! N! W =! mg θ mg cosq mg sinq θ mg F - mg cosq = N mg sinq = ma 0 F == mg cosq N a g sinq

14 ACT: Incline Consider two identical blocks, one resting on a flat surface (case A) and the other resting on an incline (case B). For which case is the normal force greater? a) case A b) case B c) both the same (N = mg) d) both the same (0 < N < mg) e) both the same (N = 0) In case A, N = W. In case B, due to the angle of y the incline, N < W. In fact, N = W cos θ. N f x θ W θ W y

15 When an object is in contact with a surface, the component of the contact force that is parallel to the surface and opposes motion is called the frictional force. Frictional Forces

16 Static Friction When the two surfaces are not sliding across one another the friction is called static friction. f f MAX s s MAX s f = µ s F N

17 Kinetic Friction Once there is motion: Kinetic friction opposes the relative sliding motion motions that actually does occur. f k = µ k F N µ k < µ s

18 ACT: Going sledding Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? a) pushing her from behind b) pulling her from the front c) both are equivalent In case 1, the force F is pushing down (in addition to mg), so the normal force is increased. In case 2, the force F is pulling up, against gravity, so the normal force is lessened. The frictional force is proportional to the normal force. 2 1

19 Inclined plane with friction F! N F N! W =! mg Increasing θ, when will the block start to slide down the plane? θ mg cosq mg sinq θ mg F N - mg cosq = mg sinq - f MAX S 0 = 0

20 What is the block s acceleration? F! N F N! W =! mg θ mg cosq θ mg mg sinq F N - mg cosq = 0 mg sinq - f k = ma

21 ACT: Sliding box A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why? a) component of the gravity force parallel to the plane increased b) coefficient of static friction decreased c) normal force exerted by the board decreased d) both a) and c) e) all of a), b), and c) As the angle increases, the component of weight parallel to the plane increases and the component perpendicular to the plane decreases (and so does the normal force). Because friction depends on normal force, we see that the friction force gets smaller and the force pulling the box down the plane gets bigger. Static friction Weight Normal

22 Example An ice skater is gliding across the ice with an initial velocity of +6.3 m/s. The coefficient of kinetic friction between the ice and the skate blades is 0.081, and air resistance is negligible. How much time elapses before his velocity is reduced to +2.8 m/s? v 0 = +6.3 m/s v = +2.8 m/s Δx =? a =? Δt =? f k = -µ k F N = -µ k mg From Newton s Second Law: ma = µ k mg a = µ k g v = v 0 + aδt Δt = (v v 0 )/( µ k g) =(2.8 m/s 6.3 m/s)/( (0.081)(9.80 m/s 2 )) Δt = 4.4 s