- #1

Aninnymoose

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## Homework Statement

α=30°An object ofm=10 kgis hung by two cablesα=30°andβ=45°. The tensions in the cables are___and___, respectively.

β=45°

*m*=10kg

*a*=0

*Tα=unknown*

Tβ=unknown

Tβ=unknown

THE ANSWERS SHOULD BE Ta=72N, Tβ=88N

## Homework Equations

*mg*=10kg*9.8m/s

^{2}=98N

*∑Fx=ma(x)*=0

*∑Fy=ma(y)*=0

A Free Body Diagram would show three forces: 98N facing downwards, Ta facing 30° up and to the left, and Tb facing facing 45° up and to the right.

## The Attempt at a Solution

I tried to isolate the horizontal and vertical components of the forces with

Tαcos30°+Tβcos45°=0→Tαcos30°=-Tβcos45°

Tαsin30°+Tβsin45°-98N=0→Tαsin30°+Tβsin45°=98N

I then tried to solve for T with

Tα=-Tβcos45°/cos30°

And substitute that into the vertical component equation

(-Tβcos45°/cos30°)sin30°+Tβsin45°=98N

But I seem to be walking in circles when I attempt to go any further.