Two parallel bars are spaced 10 m apart. A rope tied to one and draped over the other is holding up a 5 kg?

Question

Two parallel bars are spaced 10 m apart. A rope tied to one and draped over the other is holding up a 5 kg weight, which hangs between the bars. How much downward force must be applied to the loose end of the rope so that the section of rope between the weight and each bar is inclined at 45 degrees? Is more force required to make the angle shallower?

The rope in question is one of the special weightless ones.

Answer 1

Taking summation of horizontal forces,

T1(cos 45) = T2(cos 45)

where

T1 = tension of the rope tied to the left bar

T2 = tension of the rope draped to the other bar (where the force is applied)

Therefore, T1 = T2

Taking summation of vertical forces,

T1(sin 45) + T2(sin 45) = 5(9.8)

since T1 = T2,

2(T2)(sin 45) = 49

Solving for T2,

T2 = 34.65 N

NOTE that a heavier force will make the angle shallower.