(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A hanging weight with a mass of 25kg is suspended from a rope from a wall. The rope with the weight is wrapped around a pully which is also attached to a different wall. The angle that rope a(rope with mass) makes with the wall is 41 degrees, and the angle of the other rope with the other wall is 69.5 degrees. Find the force of tension in each rope.

2. Relevant equations

EFx=0

Efy=o

3. The attempt at a solution

EFx=0

Acos(49)-Bcos(20.5)=0

A=Bcos(20.5)/cos(49)

EFy=0

Asin(49)+Bsin(20.5)=mg

(Bcos(20.5)/cos(49))sin(49)+Bsin(20.5)=mg

B=172N

Substitute B back into A

A=172cos(20.5)/cos(49)

A=246

I got the force of tension in A to be 246N and the tension in B to be 172N. Is this correct or am I making this to easy? Also is it just a conicedence that the tension on the rope with the weight is the force of gravity?

Thanks for your help.

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# Homework Help: Rope Tension in Static Equilibrium

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