Pulleys Dynamics: Solving a Homework Statement

[Thomas2054]

Homework Statement

Not sure why, but I opened an old textbook (Dynamics - Engineering Mechanics, Meriam, 1980) and started paging through. I decided to try this problem (attached jpg) and got stuck.

The problem asks for the acceleration of the center of mass of the system of 4 weights, with the cables and pulleys being mass-less and no friction.

I started by trying to determine what applied forces I would put in place of the given 100N and 50N that would lead to a static situation. That caused a puzzling situation where if I worked from left to right I would have mass 1 supported by Ta = Tb = 1/2mg = 2.5g. That means that Tc is 2.5g. The downward force on mass 2 would be mg = 5g pulse Tc. But if mass 2 is supposed to be static I have Tc = 5g + Tf, requiring Tf = -2.5g, a negative tension in a cable!

It almost appears that there is not a static solution for the configuration as shown, and I need to solve it dynamically. Been too long and I am stuck.

It got my noodle going. I would appreciate any help.

Thanks.

 

[tiny-tim]

Hi Thomas2054!

Just add all the external forces (including the two strings to the ceiling), and divide by the total mass.

(and there can't be a static solution … the weights on the right outweigh those on the left!)

 

[Thomas2054]

When you say add all the external forces, including the two strings to the ceiling, I assume you mean that the sum of the upward forces is 100N + 50N + 50N + Tleft, where the second 50N is from the fact that the anchor tension must be the same as the applied 50N. How do you determine Tleft, i.e., the other anchor tension, the one attached to the fixed support on the left?

From that I would subtract 4 * m * g = 20g and divide the whole smash by 20 kg.

Is that correct?

What answer do you get?

 

[tiny-tim]

hmm … I'm having doubts now …

I'm trying to convince myself that T_b = T_c = 50, otherwise the left pulley will have infinite acceleration

Perhaps it's safest to do F = ma for each of the weights separately.

 

[Thomas2054]

I am willing to further discuss and perhaps solve it together. Game?

 

[tiny-tim]

(just got up :zzz: …)

it's your problem …

you start ​

 

[Thomas2054]

Fair enough.

I have tried a number of ideas and seem to end up with more variables than equations each time. The one item that vexes me is that I cannot figure out what Ta is. The tensions Ta, Tb and Tc are all the same, but with the steadily applied 100N force I am stuck on how to determine it.

This was based on your suggestion of trying to determine the acceleration for each weight individually.

Any suggestions?

 

[tiny-tim]

Hi Thomas2054!

The one item that vexes me is that I cannot figure out what Ta is. The tensions Ta, Tb and Tc are all the same …

T_a (= T_b = T_c) appears in the F = ma equations for both mass 1 and mass 2, so you should be able to eliminate it from them.

 

[Thomas2054]

hmm … I'm having doubts now …

I'm trying to convince myself that T_b = T_c = 50, otherwise the left pulley will have infinite acceleration

Perhaps it's safest to do F = ma for each of the weights separately.

I believe you are wise beyond your years! I went back to this and tried it and it works! The problem is simpler than I was imagining.

Using Ta = Tb = Tc = 50 and then just summing from left to right:

Sum of external forces = 50 + 100 + 50 + 50 - 20(9.81) = 53.8N. The second 50N on the LHS is for the leftmost anchor.

This all equals 20 * accel of mass center; accel of mass center = 2.69 m/s/s

All the author has is the answer and it matches.

Thanks.