# Evaluating tension on a string to which a pulley is attached [closed]

Suppose a pulley is attached to a string. Now the pulley itself has weights suspended on both of its sides having masses $m_1$ and $m_2$. Lets consider $2$ cases

Case 1

$m_1=2$
$m_2=3$

Case 2

$m_1=1$
$m_2=4$

Now surely in both cases $m_2$ will go down and $m_1$ will go up with different acceleration.

So my question is for both the cases, will the tension of the string to which the pulley itself is attach, be same?

Logically it should be according to me, but the textbook says it is different in both cases.

Can anyone please clear my doubt it would be even more better if an example is provided.

## closed as off-topic by AccidentalFourierTransform, Jon Custer, Yashas, Qmechanic♦Aug 17 '17 at 11:29

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## 3 Answers

So just to introduce some different terms to keep it clear which we're talking about: you've got a platform suspended by a "rope" and then on this platform you've mounted a pulley which holds a "cord" which connects two masses.

The cord moves freely through the pulley without stretching so it must have the same tension $T$ on either side. If the cord hangs straight down, these two tension forces both point in the same direction, down.

The force diagram for the platform then says that the force of the rope pulling upwards equals $m_\text{platform}~g + 2T,$ otherwise the platform would be accelerating downwards. (This is not Newton's third law, it is a statement that the platform is not accelerating and therefore by the second law the forces must balance to make the net force zero.) So indirectly the only question we really have is, which situation has the lower $T$ in the cord? That situation will also have a lower tension in the rope -- it's maybe not immediately obvious that that's where to start, but after looking at it for a second, I hope you can agree that that's where the force balance on the platform leaves us.

Now if you have two imbalanced masses on the cord, what determines the tension on the cord? Well, if the cord is not stretching then the total cord length is fixed and this means that their velocities are equal-and-opposite, $v_1 = -v_2.$ If they were not equal-and-opposite then the total length of the rope would be changing. If those velocities change over a short time then preserving this relation demands the accelerations also be equal-and-opposite, $a_1 = -a_2.$ But the accelerations in this case are just the net forces divided by the masses,$$(T - m_1~g)/m_1 = -(T - m_2~g)/m_2.$$ Collecting like terms gives $$\left(\frac1{m_1} + \frac1{m_2}\right)~T = 2~g.$$ So the tension has an inverse relationship with this factor $(1/m_1) + (1/m_2),$ and the tension is lower when this number is greater.

Hope that helps!

The tension in the rope is simply the amount of force required to accelerate $m1$ against the force of gravity.

$T = m_1(a + g)$

• No i am saying about the tension of the string to which the pulley is attached not the strings of the pulley itself – user7635691 Aug 16 '17 at 14:25

The formula for calculating tension on two masses hanging between a pulley is: $$T=\biggl({2m_1 m_2 \over m_1 + m_2}\biggr)g$$

To derive this equation, we start by understanding that the two masses attached by a massless rope with a pulley between them have the net forces of: $$F_1 = m_1 a_1 = T - m_1 g$$ $$F_2 = m_2 a_2 = T-m_2 g$$ and knowing that their accelerations are equal in magnitude but opposite in direction (one mass accelerates upwards at the same rate as the other accelerates downwards). Mathematically: $$a_1 = -a_2$$ Using this fact, we can substitute $-a_1$ for $a_2$ in our second force equation, then multiply by $-1$ to get $$m_2 a_1 = -T + m_2 g$$ Solving for T in our first force equation then substituting into this, we now have $$m_2 a_1 = -(m_1 a_1 + m_1 g) + m_2 g$$ which we can use to solve for $a_1$: $$a_1 = \biggl({m_2 - m_1 \over m_1 + m_2}\biggr)g$$ Now we can substitute this equation in for $a_1$ in the first force equation and solve for T.

$$m_1g\biggl({m_2 - m_1\over m_1 + m_2}\biggr)=T-m_1g$$

$$T=\biggl({2m_1 m_2 \over m_1 + m_2}\biggr)g$$

(Hint: to do the math yourself, separate the $m_1 g$ out and substitute ${m_1 + m_2 \over m_1 + m_2}$ for 1)