PHYSICS REVIEW PROBLEM-Confused please help!

Question

A 18kg sign is held up by the two ropes that pull on each side of the sign at 58' from the vertical direction. What is the tension on each rope?

Answer 1

I think you are talking about this problem:

Each tension force can be broken up into vertical and horizontal components like this:

You can use the trig ratios of sine and cosine to solve the right triangle:

sin(58°) = opposite/hypotenuse = T_X/T so:

T_X = Tsin(58°)

cos(58°) = adjacent/hypotenuse = T_Y/T so:

T_Y = Tcos(58°)

The two T_Y forces pull up on the sign and the weight of it pulls down. The weight (W) equals its mass times "g" (the acceleration due to gravity which is 9.8 m/s²). There is no acceleration happening (or no motion at all) in the y-direction so ∑F_y = zero therefore:

T_Y1 + T_Y2 - W = 0 and since T_Y1 and T_Y2 are the same then:

2T_Y - W = 0

2T_Y - mg = 0

2Tcos(58°) - (18)(9.8) = 0

2Tcos(58°) = (18)(9.8)

Tcos(58°) = (18)(9.8)/2

T = (18)(9.8)/(2•cos(58°))

T = 166.44 N

It would be appropriate to round to 2 sig figs so T = 170 N