Recent content by Eugen

Well then. The force acting on the pulley is F. F = m2a + m2g + m1g - m1a F = g(m2 + m1) + a(m2 - m1) F = gm + a(m2 - m1) From the equation m2a + m2g = m1g - m1a we find that m1 - m2 = xm/l, so m2 - m1 = - xm/l Substituting, we find that F = gm(1 - x2/l2) Thank you, haruspex.

I think the tension acts on both left and right. The total force should be 2T. Not quite sure, though.

I think the forces acting on the two parts of the chain are weight and tension: T - m2g = m2a m1g - T = m1a a = g(m1 - m2)/m It can be proven that (m1 - m2)/m = x/l, so that a = gx/l. As for point b), I'm still clueless.

Homework Statement On a pulley with a very small radius and negligible inertia, that rotates without friction around its fixed horizontal axis, there is a chain of mass m and length l. The chain starts sliding from its equilibrium position. Let x be the distance between the ends of the chain...

Oh. Now I get it. The force acting on m2 is the resultant of tension and friction, the friction being in the positive direction: -m3F/(m1 + m2 + m3) + m2gμ = m2F/(m1 + m2 + m3) Thank you.

The acceleration of the system is F/(m1 + m2 + m3). This means that horizontally, in the positive direction, on m2 acts the force: F' = m2 F/(m1 + m2 + m3) The tension acts horizontally in the negative direction: T = - m3F/(m1 + m2 + m3) So, on m2 acts horizontally this force: Fr = F(m2 -...

Homework Statement A sledge of mass m1 is pulled horizontally with a force F. On the sledge there is a body of mass m2 that can slide on the horizontal platform of the sledge with the friction coefficient μ. Another sledge of mass m3 is tied with a horizontal string of the body m2. Between the...

You are right, this explanation is consistent with all the problem data. But besides knowing some physics, one has to be a bit of Sherlock Holmes to solve it... :smile:

Point b of the problem text: "Give a detailed physical explanation for the fact that the tension at the point B is independent of length l." It can't be suspended above the peg if it is to hang (at least initially) from a string of length l. (that's what I understand if it is to change radius...

Assuming that the ball is suspended from the little black peg, the angle it makes in point A with the vertical is 90°. So applying my calculations, the tension in point B would be 3mg, just as specified in the problem text. But, once it arrives in point B, how could it change the radius while...

The ball is suspended from a string tied to the roof. There's only one string, so only one radius, L. l is the horizontal distance between point A and the radius in point B. Or so I interpret the image + the problem text. Unfortunately, the image is blurry.

I think the problem wants to show that T = 3mg no matter at what angle the ball begins to move. L.E: thinking about what drvrm said, I guess it is not possible that T is independent of θ. So, is the problem text wrong? (l is the horizontal line between point A and the string when the ball is in...

Homework Statement A ball of mass m is attached to a string of length L and released from rest at point A. Show that the tension in the string when the ball reaches point B is 3mg, independent of the length l. (there is an image in attachment ) [/B] Homework Equations K = mv2/2 U = mgh Fcp =...

Oh, no, I don't have any formal training in physics (or math). :)) I'm just trying to resolve a physics problem and I always get stuck in trigonometry equations. So it's not schoolwork I guess.

I'm trying to solve a physics problem which asks the value of an angle. The equation is this: sin θ / ( -2cos 20 - cos θ) = tan 20