Equilibrium from hanging weight and horizontal force

Question

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!

Answer 1

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.

Answer 2

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 $T$ $$T=\frac{F}{cos(\theta)}$$