# Thread:

1. ## Physics easy torque question

A beam is held in place by two wires, one at each end. A monkey with a mass of 50kg is hanging somewhere on the beam. There is a force of 180 N on one of the wires. Draw the picture, find the force on the 2nd wire and where the monkey is hanging if the beam is 2m long.

Any insight? T = F(d) will be used I am guessing and so far I have a line drawn labeled 2m with 2 vertical lines on the ends, and an arrow on the left vertical line to show the 180 N of force. Thanks in advance.

2. Well assuming that beam is in equilibrium, there must be zero moments, and also zero total force in all directions. In this case there is only forces in 1 direction, the vertical.

So, yes. You have a beam which is 2m long. Let's imagine that the far left is A, and the far right is B. So length AB = 2m.

Imagine that the wire at B is the one with a force of 180N.

Then the wire at A will have an upwards force, $F_A$ which creates a torque $T = -F_{A}(2)$ about point B. This torque will be clockwise, and that's why it's negative.

Then there's the monkey. The monkey has mass 50kg, and hence it has a weight $F_{monkey} = 9.81(50)$.

The weight of the monkey will cause an anticlockwise torque about B, given by $T_{monkey} = 9.81(50) \times d$ where d is the distance between the monkey and B.

Now we know that the sum of the moments about B must be zero, hence:

$\Sigma T = -2F_{A} + 9.81(50)d = 0$

And we also know that the sum of the forces is zero, hence:

$\Sigma F_{y} = 180 + F_{A} - 50(9.81) = 0$

Using the last equation, you can get a value for $F_{A}$, and if you substitute that value into the 2nd to last equation, you can get a value for $d$, and that's everything!

One major assumption here is that the beam itself has zero/negligible weight.

3. Excellent, thanks.
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