Calculating Forces in Equilibrium for Suspended Sign

[mandy9008]

Homework Statement

A 580 N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 80-N rod as indicated in the figure below. The left end the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0° angle with the vertical.

(a) Find the tension, T, in the cable.
(b) Find the horizontal and vertical components of force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions.)

The Attempt at a Solution

a. (580N + 80N)(3.0m) = T cos(30) (6.0m)
T= (660N)(3.0m)/(6.0m cos 30)
T=381.1N

b. Tv=T cos 30
Tv=381.1 N cos 30
Tv=330.0 N up

Fv= Tv-W
Fv= -(330.0 N - 660 N)
Fv=330 N up

Th=T sin 30
Th=381.1 N sin 30
Th= 190.6 N to left (-190.6N)

Fh=190.6 to the right

 

[ehild]

The Attempt at a Solution

a. (580N + 80N)(3.0m) = T cos(30) (6.0m)

You made a mistake when calculating the torque from the sign. The distance of its CM is not 3 m from the hinge.

ehild

 

[mandy9008]

so it the full 6m?

 

[ehild]

No. Where is the CM of the sign? The weight of the sign, a vertical force, attacks at the CM. What is the distance of this vertical line from the hinge?

ehild

 

[mandy9008]

so the weight of the sign is 59.2kg
how does this come into play?