Question is below please answer asap

Question

A 500 N weight is hung at the middle of a rope attached to two buildings at the same level. The cable will break if the tension exceeds 1800 N, what is the minimum angle the rope can make with the horizontal?

Answer 1

Force equilibrium requires that all forces in the vertical direction cancel out.

 

T = tension applied by each of the 2 cables at an angle of θ from the horizontal.  Set T = T_max = 1800 N.

W = weight = 500 N

 

Then

 

2(Tsinθ) = W

 

θ = sin^-1[W/(2T)] = sin^-1[500/(2*1800)] ≅ 7.98º 

Answer 2

Each wall must pull up on the cable by half the 500N downward Force

for the sum of vertical forces to be zero. If this were not true, it

would violate Newton's 2nd Law that F=ma. Since everything is

stationary, if sum of F = 0 then a=0.

Making a vector diagram we'd have sin x = 250/1800

where 1800 is the tension in the cable.

Thus x = 7.64 degrees

Correction x = 7.98 degrees

I wish I could provide a diagram of the vectors for clarity.

I do not know how to upload a jpeg for these answers.