Please HELP ME (PHYSICS) : finding the tension in a uniform bar?

Question

i have this problem: i actually solved the first part, which is finding the position of W1, but i don't know how to find for the tension string..

here's the problem:

A uniform bar weighing 60N has a length of 4m and is supported at its ends by strings. A load W= 200N is placed at some point on the bar such that it is in equilibrium when the tension in the string at the left end is 80N. (a) Locate the position of W1, and (b) find the tension in the string at the right end.

hope i get some right answers for this.. thanks!!

Answer 1

Let:

T = tension on the right string

x = distance of the load W from the left string

b. Summation of vertical forces equals zero:

T + 80 -60 - 200 = 0

T = 260 -80 = 180 N

a. Summation of moments at the left end is equal to zero:

4T - 2(60) - 200x =0

200x = 4(180) - 120 = 600

x =600/200 = 3 meters

Answer 2

You mentioned that you solved (a), the location of W1, but I will repeat the answer here anyway.

Because this is a static problem - the sum of all forces and moments must cancel each other out (ie, all the 'postive' forces/moments must be equal to the 'negative' forces/moments)

a) to solve for the location of W1 - you have to sum the moments about the right-hand string.

Because you take your moment point exactly at the right hand string, the force there does not come into effect.  Therefore, the sum of the moments is

60(2)+200(x) = 80(4)

200x=320-120

x=1m

Therefore the load W1 is located 1 m from the right hand end.

b) The second static equation, 'all forces must equal zero' for a body to  remain at rest, reveals that the sum of the forces in the strings must equal the sum of the weight and the load:

R+80=200+60

R= 180 N

The tension in the right hand string must be 180 N.