# Homework Help: Tension in Cables of Differing Angles

1. Mar 1, 2012

### Aninnymoose

1. The problem statement, all variables and given/known data
α=30°
β=45°
m=10kg
a=0
Tα=unknown
Tβ=unknown

THE ANSWERS SHOULD BE Ta=72N, Tβ=88N

2. Relevant equations
mg=10kg*9.8m/s2=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.

3. 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.

2. Mar 1, 2012

### ehild

The x components of the tension forces have opposite signs. Write

Tαcos30°-Tβcos45°=0

Correct.

That is two equation with two unknown, just go ahead. Why do you think that you move in circles?

ehild