Wrecking Ball Torque Problem

[BMK]

1. 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.


2. Homework Equations
$$\Sigma$$$$T$$= Fr = 0
Fx = -Ftcos + Rx = 0
Fy = -Wball - Wboom +Ftsin+Ry = 0

3. 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

 

[kkrizka]

One problem I see with you solution is that you used 3600 (L/2) to get the torque from the gravity on the boom. However, the force of gravity points straight down, while the L vector point at an angle of 48 to the horizontal.

 

[BMK]

So should it be 3600 sin 48 (L/2)?

 

[kkrizka]

yeah, sounds about right