Torque & Force Homework Solution

[BMK]

Homework Statement

A wrecking ball (weight = 4800 N) is supported by a boom, which may be assumed to be uniform and has a weight of 3600 N. A support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is 32, and the angle between the boom and the horizontal is 48. Find (a) the tension in the support cable and (b) the magnitude of the force exerted on the lower end of the boom by hinge at point P.

Homework Equations

$$\Sigma$$$$T$$ = Fr = 0
$$\Sigma$$Fx = -$$T$$cos$$\Theta$$ + Rx = 0
$$\Sigma$$Fy = -Wball - Wboom +$$T$$sin$$\Theta$$+Ry = 0

The Attempt at a Solution

$$\Sigma$$$$T$$ = 3600 (L/2) + 4800 L - Ft sin32 L = 0
$$\Sigma$$$$T$$ = 1800 L + 4800 L = Ftsin32L
$$\Sigma$$$$T$$ = 6600 = Ft sin 32
Ft = 12455N

Fx = -Ftcos32 + Rx = 0
Fx = -12455 cos 32= -Rx
Rx = 10562 N

Fy = -4800 - 3600sin48 + Ftsin 32 + Ry = 0
Fy = -4800 -3600sin48 + 12455sin 32 +Ry = 0
Ry = 875N

 

[BMK]

My physics teacher gave us a set of bonus problems over break and I cannot figure this one out. I got an answer but I do not think it is right. If anyone knows how to do this, could you please help me out. Thank you very much

 

[Moetasim]

(a) The tension in the support cable is 12455 N.
(b) The magnitude of the force exerted on the lower end of the boom by the hinge at point P is 10562 N.