Physics 8 Wednesday, October 19, 2011 Troublesome questions for HW4 (5 or more people got 0 or 1 points on them): 1, 14, 15, 16, 17, 18, 19. Yikes!
Troublesome HW4 questions 1. Two objects of inertias m 1 and m 2 start from rest and then interact only with each other in some way. (a) What is the ratio of their relative velocities at any time? (b) What is the ratio of their kinetic energies at any time?
Troublesome HW4 questions Problem 1 (a) Since the two objects interact only with each other, they form an isolated system. So momentum is conserved. Since the two objects start from rest (so momentum is initially zero), the system s momentum is zero at all times. Thus, at any time, m 1 v 1x + m 2 v 2x = 0. So v 1x v 2x = m 2 m 1 (b) The ratio of kinetic energies is K 1 K 2 = 1 2 m 1v 2 1 1 2 m 2v 2 2 = m 1 m 2 ( v1 v 2 ) 2 = m 1 m 2 ( m ) 2 2 = m 2 m 1 m 1
Troublesome HW4 questions 14. A worker pushes boxes in a factory. In each case decide which force has the greater magnitude the force exerted by the worker on the box or the force exerted by the box on the worker. (14a) The box is heavy and does not move no matter how hard she pushes. (A) Force exerted by worker on box is larger (B) Force exerted by box on worker is larger (C) The two forces have the same magnitude
Troublesome HW4 questions 14. A worker pushes boxes in a factory. In each case decide which force has the greater magnitude the force exerted by the worker on the box or the force exerted by the box on the worker. (14b) Some contents are removed, and now when pushed the box slides across the floor at constant speed. (A) Force exerted by worker on box is larger (B) Force exerted by box on worker is larger (C) The two forces have the same magnitude
Troublesome HW4 questions 14. A worker pushes boxes in a factory. In each case decide which force has the greater magnitude the force exerted by the worker on the box or the force exerted by the box on the worker. (14c) The worker pushes harder, and the box accelerates. (A) Force exerted by worker on box is larger (B) Force exerted by box on worker is larger (C) The two forces have the same magnitude
Troublesome HW4 questions Problem 14 In all three cases, the force exerted by the worker on the box is equal in magnitude to the force exerted by the box on the worker.
Troublesome HW4 questions 15. A pitcher has thrown a fastball toward home plate. (15a) When it is halfway to the plate, does the ball still feel the pitcher s push? Explain your answer. (15b) What forces does the ball feel, if any? (A) The ball still feels a force from the pitcher when it is halfway to home plate. So the relevant forces are the pitch (forward), gravity (downward), and air resistance (backward). (B) Once the ball leaves the pitcher s hand, it no longer feels a force from the pitcher. So the relevant forces are gravity (downward) and air resistance (backward). (C) Neither answer is correct.
Troublesome HW4 questions Problem 15 (a) The force between the ball and the pitcher is a contact force. The ball no longer feels the pitcher s push once the ball has left the pitcher s hand. (b) The ball feels the downward force of the earth s gravity and the backward force of air resistance. It s OK if you forgot to mention air resistance, since we ve neglected it so far in the course.
Troublesome HW4 questions 16. I push on a refrigerator (horizontally), but it does not move. Explain how this can be. (A) the downward force of gravity on the fridge is larger than the force that I can exert on the fridge. (B) the frictional force of the floor on the fridge exactly cancels the force of my push on the fridge. (C) the force I exert on the fridge is smaller than the force of friction that the fridge exerts on me. (D) the combined force of the fridge s weight and the friction between fridge and floor is larger than the force of my push. (E) the force exerted by the fridge on me equals the force exerted by me on the fridge, so the net force is zero. (F) the fridge pushes back on me harder than I push on it. (G) the fridge s very large inertia causes it not to accelerate.
Troublesome HW4 questions Problem 16 The refrigerator is pushed forward by me and is pushed backward by its contact with the floor. If these two forces cancel (as they will until I push hard enough to start the refrigerator sliding), then the refrigerator will remain at zero velocity. (There is an interesting fact about friction that it always takes a larger force to start an object sliding initially than to keep it sliding once it is moving. We ll study this in some detail soon.)
17. You are in a stationary elevator, so that the contact force exerted by the floor on you is equal in magnitude to the force of gravity acting on you. When the elevator starts downward, which force changes? What happens to its magnitude? (A) The contact force exerted by the floor on me increases in magnitude when the elevator starts downward. (B) Both the contact force and the gravitational force must be reduced in magnitude, so that they can remain equal and opposite. (C) Since the elevator is accelerating downward, the force I exert on the elevator becomes larger. (D) When the elevator starts downward, the force of gravity on me does not change, but the upward contact force exerted by the floor on me decreases in magnitude. (E) When the elevator starts downward, the force of gravity increases in magnitude.
Troublesome HW4 questions Problem 17 The downward force of gravity on me does not change when the elevator starts downward. The upward contact force between the floor and my feet is reduced when the elevator starts downward. Since the two forces no longer balance, my body accelerates downward.
Troublesome HW4 questions 18. When you are standing motionless on the ground, your feet are exerting a force on Earth. Why doesn t Earth move away from you? (A) Earth is so massive that even though I am pushing on it, its acceleration is imperceptible. (B) The ground is so firm that it pushes back up on me harder than I push down on it. (C) Earth pushes back on me just as hard as I push on Earth, so there is no acceleration. (D) Earth doesn t move away from me because gravity is pulling me down toward Earth. (E) Gravity is pulling Earth up toward me with a force that exactly cancels the downward force of my feet on the ground. (F) Earth doesn t move because Earth s gravitational pull on me is far greater than the push of my feet on the ground.
Troublesome HW4 questions Problem 18 Since we want to know whether Earth is accelerating, let s try to figure out the forces acting on Earth. Earth exerts a downward gravitational force on me. So I must exert an upward gravitational force (of equal magnitude) on Earth. My feet exert a downward force on Earth. That downward force on Earth is balanced by the upward gravitational force that I exert on Earth. So my standing on Earth does not cause Earth to accelerate. (Whew!)
Troublesome HW4 questions 19. You push on a crate, and it starts to move but you don t. Draw a free-body diagram for you and one for the crate. Then use the diagrams and Newton s third law of motion to explain why the crate moves but you don t. (A) The force I exert on the crate is larger than the force the crate exerts on me. (B) The crate s force on me is equal and opposite to my force on the crate. The frictional force between my shoes and the floor is equal in magnitude to the crate s push on me, while the frictional force between the crate and the floor is smaller than my push on the crate. (C) The crate and I exert equal and opposite forces on each other, but I don t move because I am much more massive than the crate.
Troublesome HW4 questions Problem 19 If the crate and I were both standing on an ice rink, then it seems clear that we would both start to move. If the crate and I were both bolted to the floor, then it seems clear that neither one of us would start to move. So the grip of the floor s friction on my feet must be greater in magnitude than the grip of the floor s friction on the crate. Let s say that I push to the right on the crate with a force F me,crate, so the crate pushes to the left on me with a force F crate,me = F me,crate. Meanwhile, the floor pushes to the right on me with a force F floor,me, and the floor pushes (by a smaller amount) to the left on the crate with a force F floor,crate. It is reasonable that F floor,crate < F floor,me, because the bottom of the crate is wood, while the soles of my shoes are rubber.
Troublesome HW4 questions Problem 19
Chapter 10 (second half) motion in a plane Reading Question #1: Can an object be accelerated without changing its kinetic energy?
Related question (part 1) Can an object moving in only one dimension be accelerated without changing its speed? (A) Yes. (B) No. (C) Not instantaneously, no. But over a finite time interval you could do something really tricky, like changing +v into v.
Related question (part 2) Can an object moving in two dimensions be accelerated without changing its speed? (A) Not instantaneously, no. (Like the last question.) (B) Yes. Over a finite time interval, you can do something tricky like + v v. And even instantaneously, yes, if the acceleration vector is perpendicular to the velocity vector. (C) No, never. Period.
Chapter 10 (second half) motion in a plane Reading Question #1: Can an object be accelerated without changing its kinetic energy? Answer: Yes. You can change an object s direction without changing its speed. So its velocity can change without changing its kinetic energy. Over a finite time interval, this is easy to arrange. Over an infinitessimal time interval, if the acceleration vector is perpendicular to the velocity vector, then direction changes, but speed does not. This will be important next week!
Reading Question #2 Suppose that I walk 3 miles east and then 4 miles north. How far am I from where I started? If you wanted to tell a pilot to fly directly from your starting point to your ending point, what compass direction would you tell her to fly? (Tell her something like, Your heading should be such-and-such degrees east of north. ) Answer: I am 5 miles from where I started. I ask the pilot to fly 37 east of north (or 53 north of east).
Summary of Chapter 10 results (vector math) Various ways to write a vector: A = (A x, A y ) = A x (1, 0) + A y (0, 1) = A x î + A y ĵ Can separate into two vectors that add up to original, e.g. A x = A x î, Ay = A y ĵ A = A x + A y Scalar product ( dot product ) is a kind of multiplication that accounts for how well the two vectors are aligned with each other: A B = A x B x + A y B y = A B cos(θ AB )
Summary of Chapter 10 results (friction) Static friction and kinetic (sometimes called sliding ) friction: F Static µ S F Normal F Kinetic = µ K F Normal normal & tangential components are to and to surface
Summary of Chapter 10 results (work) In one dimension, we learned W = F x x F x (x) dx Sometimes the force is not parallel to the displacement. For instance, what is the work done by gravity if you slide down a hill? W = F r = F x x + F y y which in the limit of many, many infinitessimal steps becomes W = F ( r) d r = (F x (x, y) dx + F y (x, y) dy)
Suppose mass = 1 kg. x = +19 m y = 13 m ~r = (+19 m, 13 m) ~ = (0, 9.8 N) F Sometimes the force is not parallel to the displacement. For instance, what is the work done by gravity if you slide down a hill? ~ ~r = Fx x + Fy y W =F W = (0 N) (+19 m) + ( 9.8 N) ( 13 m) 130 J Same answer as W = mgh.
Let s work through some examples using friction How hard do you have to push a 1000 kg car (with brakes on, all wheels, on level ground) to get it to start to slide? Let s take µ S 1.5 for rubber on dry pavement.
Let s work through some examples using friction How hard do you have to push a 1000 kg car (with brakes on, all wheels, on level ground) to get it to start to slide? Let s take µ S 1.5 for rubber on dry pavement. F Normal = mg = 9800 N F Static µ S F N = (1.5)(9800 N) 15000 N
Let s work through some examples using friction How hard do you have to push a 1000 kg car (with brakes on, all wheels, on level ground) to get it to start to slide? Let s take µ S 1.5 for rubber on dry pavement. F Normal = mg = 9800 N F Static µ S F N = (1.5)(9800 N) 15000 N So the static friction gives out (hence car starts to slide) when your push exceeds 15000 N.
Let s work through some examples using friction How hard do you have to push a 1000 kg car (with brakes on, all wheels, on level ground) to get it to start to slide? Let s take µ S 1.5 for rubber on dry pavement. F Normal = mg = 9800 N F Static µ S F N = (1.5)(9800 N) 15000 N So the static friction gives out (hence car starts to slide) when your push exceeds 15000 N. How hard do you then have to push to keep the car sliding at constant speed? Let s take µ K 0.8 for rubber on dry pavement.
Let s work through some examples using friction How hard do you have to push a 1000 kg car (with brakes on, all wheels, on level ground) to get it to start to slide? Let s take µ S 1.5 for rubber on dry pavement. F Normal = mg = 9800 N F Static µ S F N = (1.5)(9800 N) 15000 N So the static friction gives out (hence car starts to slide) when your push exceeds 15000 N. How hard do you then have to push to keep the car sliding at constant speed? Let s take µ K 0.8 for rubber on dry pavement. F Kinetic = µ K F N = (0.8)(9800 N) 8000 N
How far does your car slide on dry, level pavement if you jam on the brakes, from 60 mph (27 m/s)?
How far does your car slide on dry, level pavement if you jam on the brakes, from 60 mph (27 m/s)? F N = mg, F K = µ K mg a = F K /m = µ K g = (0.8)(9.8 m/s 2 ) 8 m/s 2
How far does your car slide on dry, level pavement if you jam on the brakes, from 60 mph (27 m/s)? F N = mg, F K = µ K mg a = F K /m = µ K g = (0.8)(9.8 m/s 2 ) 8 m/s 2 Constant force constant acceleration from 27 m/s down to zero: x = v i 2 2a = v 2 f = v 2 i + 2ax (27 m/s)2 2 (8 m/s 2 ) 45 m
How far does your car slide on dry, level pavement if you jam on the brakes, from 60 mph (27 m/s)? F N = mg, F K = µ K mg a = F K /m = µ K g = (0.8)(9.8 m/s 2 ) 8 m/s 2 Constant force constant acceleration from 27 m/s down to zero: x = v i 2 2a = v 2 f = v 2 i + 2ax (27 m/s)2 2 (8 m/s 2 ) 45 m How much time elapses before you stop?
How far does your car slide on dry, level pavement if you jam on the brakes, from 60 mph (27 m/s)? F N = mg, F K = µ K mg a = F K /m = µ K g = (0.8)(9.8 m/s 2 ) 8 m/s 2 Constant force constant acceleration from 27 m/s down to zero: x = v i 2 2a = v 2 f = v 2 i + 2ax (27 m/s)2 2 (8 m/s 2 ) 45 m How much time elapses before you stop? v f = v i + at t = 27 m/s 8 m/s 2 = 3.4 s
How does this change if you have anti-lock brakes (or good reflexes) so that the tires never skid?
How does this change if you have anti-lock brakes (or good reflexes) so that the tires never skid? Remember µ S > µ K. For rubber on dry pavement, µ S 1.5 (though there s a wide range) and µ K 0.8. The best you can do is maximum static friction: F S µ S mg a = F S /m = µ S g = (1.5)(9.8 m/s 2 ) 15 m/s 2
How does this change if you have anti-lock brakes (or good reflexes) so that the tires never skid? Remember µ S > µ K. For rubber on dry pavement, µ S 1.5 (though there s a wide range) and µ K 0.8. The best you can do is maximum static friction: F S µ S mg a = F S /m = µ S g = (1.5)(9.8 m/s 2 ) 15 m/s 2 Constant force constant acceleration from 27 m/s down to zero: x = v i 2 2a = v 2 f = v 2 i + 2ax (27 m/s)2 2 (15 m/s 2 ) 24 m How much time elapses before you stop? v f = v i + at t = 27 m/s 15 m/s 2 = 1.8 s So you can stop in about half the time (and half the distance) if you don t let your tires skid. Or whatever µ K /µ S ratio is.
Steel on steel µ K is about half that of rubber on concrete, and much less than that of µ S for rubber on concrete. So a train can take a while to skid to a stop! Even more so if the tracks are wet: µ K 0.1 At µ = 0.1 on level ground: 360 m to stop from 60 mph. At µ = 0.1 on 6 slope: not possible to stop.
Friction on inclined plane
Take x axis to be downhill, y axis to be upward from surface. F gravity = mg cos θ ĵ, F contact = +mg cos θ ĵ F gravity = +mg sin θ î If block is not sliding then friction balances downhill gravity: F static = mg sin θ î
Magnitude of normal force ( normal is a synonym for perpendicular ) between surfaces is F N = mg cos θ Magnitude of static friction must be less than maximum: F S µ S F N = µ S mg cos θ Block begins sliding when downhill component of gravity equals maximum magnitude of static friction...
Block begins sliding when downhill component of gravity equals maximum magnitude of static friction: µ S mg cos θ = mg sin θ µ S = mg sin θ mg cos θ µ S = tan θ
A heavy crate has plastic skid plates beneath it and a tilted handle attached to one side. Which requires a smaller force to move the box? (A) Pushing the crate is easier than pulling. (B) Pulling the crate is easier than pushing. (C) There is no difference.
Which path best represents the trajectory of a cantaloupe thrown horizontally off a bridge? (What s wrong with the other two?)