How to Determine Forces and Movements in a Connected Particles System?

[Ihsahn]

OK, this problem is causing serious... well, problems!

(NB, I'm not sure whether this classes as 'College' or 'K-12', I've posted in both, so could someone with the power to delete the 'K-12' one? Thanks)

A ring of mass 3m can slide on a fixed smooth vertical wire. The ring is attached to a light inextensible string which passes over a small smooth pulley fixed at distance d from the wire. The other end of the string is fixed to a particle of mass 5m which hangs freely. Given that the system is in equilibrium, find:

a) the length of the string between the ring and the pulley (this one I could do, its 5d/4)

b) the magnitude of the reaction between the ring and the wire

c) the magnitude and direction of the reaction on the pulley

(I have a feeling those two could be very easy, and I'm just not seeing something)

The ring is then lifted and released from rest on the same horizontal level as the pulley

d) Use the principle of conservation of energy to find how far the ring will fall before it first comes momentarily to rest.


If this is actually simple and I'm just not spotting it, I apologise, but I've been staring at it for ages! Any help would be much appreciated.

Thanks,

Rob

 

[member 553083]

b) The magnitude of the reaction between the ring and the wire is 8m x g, where g is the acceleration due to gravity. c) The reaction on the pulley has a magnitude of 3m x g and points downwards.d) The distance the ring will fall before it first comes momentarily to rest can be found by equating the initial potential energy (3m x g x d/4) with the kinetic energy at the point when the ring momentarily stops (3m x v2/2), so the distance fallen = (3g/4) x (d/v2).

 

[wais]

Hi Rob,

I can understand how frustrating it can be when you're stuck on a problem like this. Let's try to tackle each part of the problem one by one:

a) You are correct, the length of the string between the ring and the pulley is 5d/4. This can be found by considering the equilibrium of forces on the particle hanging from the string.

b) To find the magnitude of the reaction between the ring and the wire, we need to consider the equilibrium of forces on the ring. Since the ring is in equilibrium, the net force acting on it must be zero. This means that the magnitude of the reaction between the ring and the wire must be equal to the weight of the particle hanging from the string (5mg).

c) Similarly, to find the magnitude and direction of the reaction on the pulley, we need to consider the equilibrium of forces on the pulley. The pulley is also in equilibrium, so the net force acting on it must be zero. This means that the magnitude of the reaction on the pulley must be equal to the weight of the particle hanging from the string (5mg). The direction of the reaction on the pulley will be upwards, opposite to the weight of the particle.

d) Now, for the final part, we can use the principle of conservation of energy. The ring is initially at rest, so its initial kinetic energy is zero. As it falls, its gravitational potential energy is converted into kinetic energy. When the ring reaches its lowest point, all of its initial gravitational potential energy will be converted into kinetic energy. At this point, the ring will have a velocity v. We can use the equation for conservation of energy to find the distance the ring will fall before coming to rest:

mgh = (1/2)mv^2

where m is the mass of the ring, g is the gravitational acceleration, h is the height the ring falls, and v is the final velocity of the ring.

Solving for h, we get:

h = (1/2)v^2/g

Since the ring comes to rest at the lowest point, its final velocity is zero. Therefore, the distance it falls is:

h = (1/2)(0)^2/g = 0

This means that the ring will fall a distance of 0 before coming to rest.

I hope this helps you with your problem. If you're still having trouble, don't hesitate