# Dependent motion between 2 bodies to find velocity

#### t_n_p

1. Homework Statement

If the system is released from rest, determine the speeds of both masses after B has moved 1m. Neglect friction and the masses of the pulleys.

http://img14.imageshack.us/img14/6321/fbd.jpg [Broken]

3. The Attempt at a Solution
I have drawn up the free body diagram for both blocks, as well as labelled the lengths Sa and Sb relative to their respective datum points/lines.

I obtain the equation 3Sa + 2Sb = L* where L* is the length of rope that is constant
this leads to the velocity and acceleration equations:

3Va = -2Vb and 3Aa = -2Va

My question is where do I go from here?
Do I do newtons second law, sum of forces = ma?
This only gives me acceleration in terms of tension T

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#### rl.bhat

Homework Helper
First of all you have to find the tension in the rope.
Since rope is single, acceleration of both the masses must proportional to the change in the lengths. .
Calculate the accelerations of the both the masses and equate them. That gives you the tension. If the know the acceleration of either mass, using kinematic equation, you can find the the velocity.

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#### t_n_p

Since rope is single, acceleration of both the masses must be the same.
I have seen other examples with 2 hanging blocks suspended by a pulley system where there was only one rope and acceleration of the two blocks was different.

Additionally, what is the point of the acceleration equation 3Aa = -2Va, if it is contrary to what you say?

#### rl.bhat

Homework Helper
OK. I have edited the previous post. When B moves 1 m downward how much A moves upward?

#### t_n_p

Firstly, I would be inclinded (pardon the pun) to say block A moves downward and block B moves upwards.

Secondly, I am not 100% sure mathematically how to determine how far A would move when B moves 1m. I am tempted to say 1.5m due to their being 3*Sa and 2*Sb

#### rl.bhat

Homework Helper
OK. Now find acceleration of A and B and equate them in the ratio 2Aa = 3Bb
Take the acceleration positive in both cases i.e in the direction of increasing velocity.

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#### t_n_p

Ok, If B moves 1m up, Sb will shorten by 2m.
Hence Sa will be 2/3m as sort of suggested by the displacement equation.

Mathematically, how do we come to this conclusion from: 3Sa + 2Sb = L*

=> 3Sa = L* - 2Sb
=> Sa = (L* - 2Sb)/3

What does this mean physically?

#### rl.bhat

Homework Helper
What does this mean physically?
It means that if B moves x m, A moves 2/3 x m. Since they take same time for the displacement Bv =2/3Av and Ba = 2/3 Aa

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#### t_n_p

What does this mean physically?
It means that if B moves x m, A moves 2/3 x m. Since the take same time for the displacement Bv =2/3Av and Ba = 2/3 Aa
I could understand that if the displacement equation was Sa = 2Sb/3
I am confused due to the constant length L*

Sa = (L* - 2Sb)/3
Sa = L*/3 - 2Sb/3

Doesn't the L*/3 also affect the length of Sa?

#### rl.bhat

Homework Helper
Total length L = 2Sb + 3Sa.
Take time derivative on both side. Note that L is constant. That gives you the velocity of A and B. Again take the time derivative of velocities. That gives you the accelerations of A and B.

#### t_n_p

Ok, I found
Aa = -1.099 m/s^2
Ab = 1.649 m/s^2

I know acceleration and initial velocity and I want final velocity when B has moved 1m.

I want to use Vf^2 = Vi^2 + 2as, but acceleration isn't constant because it starts from rest?

#### rl.bhat

Homework Helper
When you drop any object from a height, it also starts from rest. Still it has a constant acceleration. Similarly here also the initial velocity is zero and acceleration is constant.

#### t_n_p

how do I prove/show that acceleration is constant?

#### rl.bhat

Homework Helper
The movements of A and B are caused by the gravitational forces on them.
When an objects slides on an inclined plane, its acceleration decreases but it remains constant.

#### t_n_p

don't you mean it constantly increases since it starts from rest?

#### rl.bhat

Homework Helper
don't you mean it constantly increases since it starts from rest?
No. The velocity increases.
When we write v = u + at, a remains constant.
When one mass is pulling other, both must have the same acceleration.

#### t_n_p

Ok, now that I have established acceleration is constant, I am able to use Vf^2 = Vi^2 + 2as.

However, since my acceleration of block a is negative, I run into a problem with taking square root of a negative. Have I done somthing wrong?

#### rl.bhat

Homework Helper
Negative or positive sign indicates the direction of the acceleration. You have to find the magnitude of the acceleration. For that take the positive value.

#### t_n_p

cool.

I figured out V of block b = 1.82m/s (upwards) and V of block a = 1.21m/s (down the incline)
These answers seem reasonable to me, I am confident they are correct.

Thanks for the help

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