Tension HELP!?

Question

Consider the 662N weight held by 2 cables... The left hand cable had tension T and makes an angle θ with the wall. The right hand cable had tension 860N and makes and angle of 32 degrees w/ the ceiling.

- What is the tension T in the left hand cable slanted at an angle θ w/ respect to the wall?


- What is the angle θ wich the left hand cable makes w/ respect to the wall?


**I understand the problem if the angles were given - but how could you find the tension w/o the angle???

Answer 1

You know the total vertical is 662 N and the horizontal and vertical components of the right hand cable, so

vertical:
662=860*sin(32)+T*cos(θ)


where T is the tension in the left hand and θ is an angle subtended by the wall and the cable below the cable.

Horizontal
860*cos(32)=T*sin(θ)

two equations and two unknowns.

662=860*sin(32)+T*cos(θ)
T=(662-860*sin(32))/cos(θ)

860*cos(32)=T*sin(θ)
T=860*cos(32)/sin(θ)

Tan(θ)=860*cos(32)/(662-860*sin(32))

θ= 74.2 degrees

T=758 N

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Answer 2

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