I didn't literally guess. I just wasn't sure of the answer, so I used the word "guess". I think the situation is similar to the one walking on ice. For every inch of rope you pull down its original position, your hand reaches 2 inches up the rope (assuming that he stretches two full arm...
I do not think the swing seat would be propelled backwards. The person on the swing has acquired his speed due to the drop of his own potential energy. But I have another question: how do we know what speed will the person leave off the swing? And I think that if man leaves off the swing with...
I meant that if the conservation of energy applies. The kinetic energy of the person and the swing seat at the bottom point would be equal to the sum of the kinetic energy of the seat and the person after he flies off the swing seat. The latter energy does not add up to the initial energy. So I...
On a standard pulley, on one end of the rope hangs a mass equal to a person's weight. That person is climbing up on the other end of the rope at constant speed. My guess is that the man and the mass will rise up at the same speed.
And does the conservation of momentum apply in this case? How...
The gravitational force on the center of mass indeed does not pass through the point of contact, but that does that produce a net torque on the ball. The normal force on the ball acts through the center of mass and therefore affect the ball uniformly. The net force on the ball drags it down the...
I think that gravity does exert a net torque on the ball. I think that normal force acts at angle on the ball, and thus will cancel out some of the gravitational forces on the ball. The front of the ball then gets more torque than the back of it.
I have posted this question before but have not got a complete answer. I have since been thinking about it quite often, yet still have not had a conclusive answer. I'd really appreciate if someone can give a full explanation, since it is a fact known by most.
1. Under all ordinary conditions...
g=9.8 ms^-2
a=2 ms^-2
u1= coefficient of friction between the two blocks
The upper pulley acc. at 1/2 of the acc. of the bottom block.
So I got 2T-2.29*g*u1=2.29*a/2
T=2.83 N
F-5.65*g*u2-T=(5.65+2.29)*a
F=20.73 N
First I thank your responses. To mgb_phys: I have googled this answer for lots of times; the only satisfactory answer came from wikianswers. I only wanted to make sure this is really the case. To Doc Al: Friction.
I guess I am going to think about this problem a bit more.
I am sorry. Can you be more specific as to which answer you responded yes? I had two situations. I only want to hear what others think so that I can verify my own answers. Thank you for responding.
1. Under all ordinary conditions, would a ball released from rest start rolling down a frictionless inclined plane with just the force of gravity pulling it down? Or is it really just going to slide down?
Now let's suppose a gigantic ball released on the slope of an enormous frictionless...