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Mechanics question involving pulleys: where is the problem?

  • Thread starter Anonymouse
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Question

A string is hung over a fixed pulley, and a 4kg mass is suspended from one end of the string. The other end of the string supports a massless pulley, over which a second string is hung. This string has a 2kg mass attached to one end, and a 1kg mass attached to the other end. See the diagram below:
View attachment pulleys2.bmp
If the system is released from rest, what are the tensions T1 and T2 in the strings? What are the accelerations of each mass, aA, aB and aC?

Solution

Firstly, let's define the upwards direction as the positive direction. Take the gravitational constant g to be 10, so the weight of mass m is given by 10m. Then if we apply Newton's 2nd law (F=ma where F is force, m is mass and a is acceleration) to each of the masses, we get three equations:

For mass A: T1 - 40 = 4aA

For mass B: T2 - 20 = 2aB

For mass C: T2 - 10 = 1aC

Now it can also be shown that 2aA + aB + aC = 0.

Now we have four equations in total, but five unknown quantities. How can we solve the problem?

I guess we could also consider masses B and C as a composite, and apply Newton's 2nd law again to arrive at the following equation:

T1 - 30 = -3aA

In this equation, there is a minus sign on the right-hand side because the acceleration of the composite of B and C must be of equal magnitude but opposite direction to that of A, in other words -aA, right?

So now we have five equation with five unknowns, which can be solved to give the following answers:

T1 = 240/7 = 34.29 N

T2 = 320/21 = 15.24 N

aA = -10/7 = -1.43 m/s²

aB = -50/21 = -2.38 m/s²

aC = 110/21 = 5.24 m/s²

Is this solution correct?

Problem

If we consider just the massless pulley, it has an upward force of T1 on it, and a downward force of 2T2.

In principle we could apply Newton's 2nd law, but since mass is zero, the net force must also be zero in order to avoid getting an infinite acceleration.

So if the net force is zero, then T1 = 2T2, but this is clearly not the case if the values calculated above are correct.

So, how can this discrepancy be accounted for?
 

Answers and Replies

  • #2
AlephZero
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I guess we could also consider masses B and C as a composite, and apply Newton's 2nd law again to arrive at the following equation:

T1 - 30 = -3aA
The acceleration of the "composite mass" is not -aA. The acceleration of the massless pulley is -aA but the CG of the "composite mass" has an acceleration relative to the pulley.

Your other equation for the equilibrium of the massless pulley, T1 = 2T2, is correct.
 
  • #3
The acceleration of the "composite mass" is not -aA. The acceleration of the massless pulley is -aA but the CG of the "composite mass" has an acceleration relative to the pulley.
Yes, that makes sense.

So the acceleration of the centre of mass of the BC composite aBC is given by 3aBC = 2aB + aC (this is just saying the rate of change of momentum of the composite is equal to sum of the rates of change of momentum of B and C separately).

So we can use Newton's 2nd law for the composite:

T1 - 30 = 3aBC = 2aB + aC

So now using the five equations, we can solve and get the following values:

T1 = 32 N

T2 = 16 N

aA = -2 m/s²

aB = -2 m/s²

aC = 6 m/s²

As can be seen, now T1 = 2T2 as the case should be.
 

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