Tension in a two pully system

  • Thread starter pinokicake
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  • #1

Homework Statement


Two masses hang by a pulley system shown in the attachment. The masses both weigh the same. Find the tension on the rope.

The first attachment is straight from the book, the second attachment is me redrawing it with labels along with a free body diagram of each mass (vectors not to scale) and the third is another picture with labeled ropes (its relevance is in the explanation of my thinking under "attempt at solution").


Homework Equations



m1 = mass on the left a1 = acceleration of m1

m2 = mass on the right a2 = acceleration of m2


m1 = m2 = m

T = tension

F = ma

Weight = mg

Ideal Mechanical Advantage (IMA) = Fout/Fin

4. The answer provided

The answer is 3/5mg. I will go into how they arrive to this in #5

4. The attempt at a solution

Alright, I know my main problem is this figuring out which mass I designate the input force and which i designate the output force.

My first attempt was saying the the input force would be where m1 hangs and the output force would be where m2 hangs. I also know that with a pulley, the mechanical advantage is the amount of strings supporting movable pulley. So with that said:

2 = Fout/Fin

2 = ma2/ma1

2 = a2/a1

2a1 = a2 this is where my mistake is, the book claims that m1 should be moving twice as fast as m2 or in other words a1 = 2a2.

Now I understand that this should be true. In the third attachment I labeled all three ropes starting with the left most rope 1, 2 and 3. If rope 1 were to lengthen a total of 2 meters, both ropes 2 and 3 need to shorten a total of 1m. This would mean that m1 would move down 2m and m2 would move up 1m. Both the masses are moving at the same time, so for each mass to move to their respected heights, m1 needs to go twice as fast as m2 (it has to cover twice as much distance as m2).

5. The worked solution

With the correction in mind the following is the worked solution to the problem:

-F1 = -ma1 = T - mg

F= ma1 = mg - T

a1 = 2a2

2ma2 = mg - T

F2 = ma2 = 2T - mg

a2 = (2T - mg)/m

2m[(2T - mg)/m] = mg - T [on the left side of the equation m in the denominator cancels and we distribute the 2]

4T - 2mg = mg -T

5T = 3mg

T = 3/5mg
 

Attachments

Answers and Replies

  • #2
ehild
Homework Helper
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What is your question?

You can use the mechanical advantage for zero acceleration. If the question was what should be m1 to balance m2, the answer would be that the output force (m2g) is twice the input force (m1g) see http://en.wikipedia.org/wiki/Mechanical_advantage

It is not the case here. The blocks accelerate - it is not balance.

Such problems are solved by assuming constant length of the rope. The length of the pieces 1, 2, 3 should remain constant. L1+L2+L3 = const. The rates of change of the lengths sum up to zero: dL1/dt +dL2/dt+dL3/dt=0. Moreover, L2=L3. If the velocity of block m1 is v1 downward it is v1= dL1/dt, the length of piece 1 increases. The velocity of m2 then should be v2=dL2/dt=dL2/dt=-0.5 v1, the block raises and pieces 2,3 shorten.

The same is true for the accelerations: a1=-2a2. Just so as you explained.

ehild
 
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  • #3
Thank you ehild! My question was pretty much why I couldn't use mechanical advantage to solve the problem, and as you pointed out you can only use mechanical advantage when there is zero acceleration.
 

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