Pulley and acceleration problem

Jan 2015
96
1
Hi everybody out there, hope you are having a great day. I have tricky problem (well I have found it tricky) and I just can't see how to arrive at the correct solution. I hope someone can give me some help in tackling the question.

The problem involves a mass (m1) which sits on a smooth flat surface, above the ground (ie. a table-top) . m1 is attached by a rope/string through two pulleys to another mass (m2). The first attachment (first pulley) is at the table-top edge at the same horizontal height as m1. Directly below the first pulley is a second pulley and the rope runs through this pulley and back up directly vertically where it is attached to a hook to the ceiling. The second mass is attached directly to the second (lower pulley) hanging beneath the pulley above the ground.
I hope my description of this system is not too confusing.

The problem asks in terms of m1, m2 and g (gravity), find the acceleration of each mass.
NOTE - there is no friction anywhere in the system.
 
Apr 2015
1,035
223
Somerset, England
Clearly the system is not in equilibrium

So what laws or analysis are you attempting.

Hint you need to calculate the net force on each of m1 and m2 and connect this to the accelerations.

So how far have you got?
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
Start by drawing a force diagram for both of the masses. And note that because of the way the pulleys are arranged the acceleration of mass 2 is 1/2 the acceleration of mass 1. What equations of motion do you come up with?
 
Apr 2015
1,035
223
Somerset, England
Hello, Jack, thank you for the email, We all hope you are fully recovered from your illness.

This type of problem is called the motion of connected bodies.
The secret is to figure out what is connected and what additional equations can be had because of this connection.

Did you draw a diagram?

With reference to mine

Mass M1 only moves horizontally and mass M2 only moves vertically.
There is no friction anywhere so the tension is constant throughout the string. Call this T.
(Generally if friction is involved there will be different tensions in different parts of the string)

So writing Newton's Law equations for the motion Force = mass times acceleration for each particle with accelerations a1 and a2.

This allows us to substitute for the tension and eliminate it, but leaves us with a1 and a2.

Now chip has told you that a1 = 2a2.
This is easy to say, but can you prove it mathematically?
(This bit is often omitted in textbooks)

Well because the string does not stretch, break or go slack, the amount of string passing over pulley1 horizontally must be the same as the total amount of string that is added to both of the two support sides of the pulley2 as it descends.
So half the string is added to each side of pulley2 support so pulley2 descends vertically half the distance m1 moves horizontally.

But both m1 and m2 start from rest so at any time t from rest
s1 = 2s2.

So resorting to our old friend s = ut + 1/2 a t^2 we can show

a1 = 2 a2 as chip said.

I will leave you to finish the algebra to obtain the correct answers.
 

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Last edited:
Jan 2015
96
1
Hi STUDIOT,

JackTheHat here,
Thank you for that explanation of what exactly was going on in the pulley system in the problem. I see it now .. It hadn't occurred to me to consider how the length of rope that was going through either side of the 2nd pulley was related to the distance that the first 1st mass moves. Although now I see it has to be crucial to solving the problem. Now that you have brought my attention to that part of the problem in addition to your excellent diagram .. I think I can complete the solution correctly. I had just got myself so mixed up that I could no longer see exactly what was happening . I had hit a brick wall.
Thank you again for taking the trouble to help. It is much appreciated.

Regards,
JackTheHat
 
Jan 2015
96
1
Hi ChipB,
I think I now have enough to get the solution to the problem, thanks to your help.
Thank you for taking the trouble to give me some hints on solving the thing, it is much appreciated.
Regards,
JackTheHat.
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
Hello JtH - glad you were able to figure it out! Good luck with your studies.
 
Oct 2017
1
0
Thanks for the Diagram.. I was overlooking the tensions on the vertical m2.