Calculating Tension and Net Force in Suspended Cable System

[Ellen W.]

Homework Statement

A 50.0 kg 6.00m awning is held up by a wire in the middle. A 15kg sign hangs at the end. Calculate the tension in the massless wire and the magnitude of the net force from the wall acting on the awning.
The cable forms a 40 degree angle with the awning at the bottom right corner.

Homework Equations

Sum or Torque

The Attempt at a Solution

Sum of Torque=0=(mg)(L)+(Mg)(L/2)-(T)(L/2)Sin40)
L's cancel
(mg)+(Mg/2)=Tsin40/2
(mg)+(Mg)/sin40=T
((15*9.81)+(50*9.81)/sin40)=T
T=992N
?
I'm not sure what the question means by net force from the wall. Is it just the normal force of the wall or do I have to use the tension to find it?

 

[TSny]

(mg)+(Mg/2)=Tsin40/2
(mg)+(Mg)/sin40=T

Careful with treating the 2's.

I'm not sure what the question means by net force from the wall. Is it just the normal force of the wall or do I have to use the tension to find it?

What can you say about the net force on the system?

 

[Ellen W.]

The net force would have to be zero. Does this mean that they are equal?

 

[TSny]

It means ∑F_x = 0 and ∑F_y = 0. The wall exerts a force on the awning that could have both and x and y components.