# Tension in a two pully system

**1. The problem statement, all variables and given/known data**

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").

**2. Relevant equations**

m_{1} = mass on the left a_{1} = acceleration of m_{1}

m_{2} = mass on the right a_{2} = acceleration of m_{2}

m_{1} = m_{2} = m

T = tension

F = ma

Weight = mg

Ideal Mechanical Advantage (IMA) = F_{out}/F_{in}

**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 m_{1} hangs and the output force would be where m_{2} hangs. I also know that with a pulley, the mechanical advantage is the amount of strings supporting movable pulley. So with that said:

2 = F_{out}/F_{in}

2 = ma_{2}/ma_{1}

2 = a_{2}/a_{1}

2a_{1} = a_{2} this is where my mistake is, the book claims that m_{1} should be moving twice as fast as m_{2} or in other words a_{1} = 2a_{2}.

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 m_{1} would move down 2m and m_{2} would move up 1m. Both the masses are moving at the same time, so for each mass to move to their respected heights, m_{1} needs to go twice as fast as m_{2} (it has to cover twice as much distance as m_{2}).

**5. The worked solution**

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

-F_{1} = -ma_{1} = T – mg

F= ma_{1} = mg – T

a_{1} = 2a_{2}

2ma_{2} = mg – T

F2 = ma_{2} = 2T – mg

a_{2} = (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

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