# Math Help - Forces and components

1. ## Forces and components

A chain of weight W is slung between two walls such that the two ends are level. Given that the chain is slack, and that the angle formed between the normal to the wall and the chain at each end is $\theta$, find the tension at the bottom of the chain.

This is part b of the same question.
Part a asks for the the tension at each end of the chain. That is obvious because the sum of the vertical components of the tension (T) at each end = weight of the chain. This implies that W=2Tsin $\theta$, hence T=W/(2sin $\theta$)

Can anyone help with part b?

2. Get a free-body diagram of half of the chain.

Cut the chain at the "bottom of the chain, at midway between the two walls.
Consider the right half only.
So that it, the cut middle point, the "bottom of the chain", will stay exactly where it was before your cutting away the left-half, it should be supported by the tension at that cut point. This is the tension that you are looking for. Let's call it tension U, as T is already taken.

Tension is pulling away. So, the horizontal component of U, Uh, is pointing to the left.
The vertical component of U, Uv, is pointing upwards.

------------------------------------
Its vertical component, Uv:

----At right end where it is attached to the right wall, the wall is pulling the chain equal to the tension, T, at that end. The vertical component of which, you got in the part a) of the Problem , is W/2.

----The weight of the half-chain is W/2 also. This weight can be imagined to be concentrated at the middle of this half-chain.

So, summation of vertical forces equals zero,
Uv -W/2 +W/2 = 0
Uv = 0
Meaning, tension U has no vertical component. U is all horizontal.

-------------------------------------
The horizontal component of tension U, Uh:

Uh = U itself.

---The vertical compopnent of T(tension at the wall suppost), Tv = W/2 = T*sin(theta)
So, W/2 = T*sin(theta)
T = W/(2sin(theta)) ------as you found out.

Hence, the horizontal component of T, Th,
Th = Tcos(theta)
Th = [W/(2sin(theta)]cos(theta)
Th = Wcos(theta) / 2sin(theta)
Th = (W/2)cos(theta) / sin(theta)
Th = (W/2)cot(theta) -------------------***

Now, in the free-body diagram, summation of horizontal forces equals zero,
-Uh +Th = 0
Uh = Th
or,
U = Th
U = (W/2)cot(theta) -----the tension at "the bottom of the chain", answer.