Last problem on math homework has me stumped for two days, I would appreciate any help on this.

Question

A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. The hoist weighs 390 N. The ropes, fastened at different heights, make angles of 50° and 38° with the horizontal. Find the tension in each rope and the magnitude of each tension. (Let T2 and T3, represent the tension vectors corresponding to the ropes of length 2 m and 3 m respectively. Round all numerical values to two decimal places.).

I have been stuck on this problem for two days. I've tried it many times and got different answers to all of the parts of the question, and every single time all of the parts were wrong. I looked online to see if there were any similar problems, and there were, so I followed the same steps they took to solve it, but my answers from doing that are also incorrect. I only have two more attempts to get it right and it's the last problem on my math homework.

Thank you so much to anyone who helps!

Answer 1

Hi Alyssa,

These problems are always easier to solve if you have a sketch. But I can't make a sketch so we will do the best we can.

First of all we know that the hoist is at rest ( no acceleration) so by Newtons 2nd law the sum of the external forces acting on it are zero. So we have T3 + T2 + W = 0 where W =390 N now this is a vector equation so we need to write it in terms of its components.

y direction: sin(38)*T3 + sin(52)*T2 -W=0

x direction:-cos(38)*T3 + cos(52)*T2=0

using sin(38)=.62, sin(52)=.79, cos(38)=.79 and cos(52)=.62 and W=390

solve the 2 equations for the 2 unknown tensions T2 and T3

I got T2=305.50 N and T3=239.76 N

Let me know if you have questions

Jim

Answer 2

Use T₂ sin50° +T₃ sin38° =390 N and T₂ cos50° = T₃ cos38°. This gives T₂ = 1.226 T₃ and .766 T₃ +.616T₂ = 390 N. using the expression for T₂ in the second equation gives T₃ = 250.84 N and gives T₂ = 307.5 N as the tensions in the cords.