# Equilibrium from hanging weight and horizontal force

Here is the problem:

A light inextensible string of length 40cm had its upper end fixed at a point A, and carries a mass of 2kg at its lower end. A horizontal force applied to the mass keeps it in equilibrium, 20cm from the vertical through A. Find the magnitude of this horizontal force and the tension in the string.

Ok so firstly I don't understand why it needs to have a horizontal force to keep it in equilibrium when it's attached to a fixed point?

I've drawn a diagram and tried to use Lami's Theorem but I've got too many variables I think and I can't see how the position of the force is going to be helpful?

I defined F as the horizontal force and $$\theta$$ as the force between the tension T and the horizontal force F. Lami's Theorem gave me:

$$\frac{2g}{sin(\theta)} = \frac{T}{sin(90)} = \frac{F}{sin(270-\theta)}$$

which gives

$$\frac{2g}{sin(\theta)} = T = \frac{F}{-cos(\theta)}$$

Any pointers greatly appreciated, thank you!

If there were no force the weight would hang straight down. When you push horizontally, you move the equilibrium to the side. You have three forces, gravity on the mass, which is downward, the horizontal force, and tension in the string, which is along the string. These have to add to zero. You can separate the two directions. The tension horizontal and vertical values are related by trig functions of the angle of the string.

Find the angle at $$A$$. The vertical component of $$T$$ has to counter gravity, which gives $$T$$. The horizontal component of $$T$$ counters the horizontal force, so you can compute the horizontal force.

• +1 This is how I'd solve the present question, but for slightly different versions of the question, the following observation might be useful, so I'll record it here: If you were asked only for the horizontal force, not for the tension in the string, you could compute the former without first computing the latter. Use that the resultant of the horizontal force and the gravitational force has to be in the direction of the string. Mar 21, 2020 at 15:36
• Oh, my diagram was wrong - I had interpreted it as the gorizontal force being 20cm down from A, Thank you! Mar 25, 2020 at 10:50

Let $$m$$ be the mass, $$\theta$$ be the angle between the tension force and the horizontal force, $$T$$ be the tension force, and $$F$$ be the horizontal force.

These problems are often easier when you split them into the horizontal and vertical components.

Since the mass is in equilibrium, $$\sum{F_{y}}=0$$ $$\implies mg=Tsin(\theta)$$ and $$\sum{F_{x}}=0$$ $$\implies F=Tcos(\theta)$$ First, find the horizontal force $$F$$ $$T=\frac{mg}{sin(\theta)}$$ $$T=\frac{F}{cos(\theta)}$$ $$\frac{mg}{sin(\theta)}=\frac{F}{cos(\theta)}$$ $$F=mgcot(\theta)$$ You can plug in your values here (calculating $$\theta$$ using the given distances)

Then, solve for $$T$$ $$T=\frac{F}{cos(\theta)}$$

• Thanks Weston - I've worked through it and it makes sense to me. How would I work out the size of $\theta$? I don't see how to use the distances Mar 23, 2020 at 14:17
• @jemima you can use the fact that for a right triangle, $cos(\theta)=\frac{adjacent}{hypotenuse},$ with the hypotenuse being the length of the string, and the adjacent side being the horizontal displacement. It might help to either draw a picture of it or reference the one in Ross's answer above. Mar 23, 2020 at 22:38
• @jemima to clarify, you do not need to necessarily calculate $\theta$ in order to solve the problem; knowing the length of the three sides of the triangle referenced in my previous comment, you can find the value of trigonometric functions (which is all you need in the problem). If, however, you did need to know $\theta,$ you could use an inverse trig function to solve for it. Mar 23, 2020 at 22:41
• Ah, my diagram was all wrong - I had misinterpreted the question. Thanks so much for your help!! Mar 25, 2020 at 10:49