How Is the Force on the Boom Hinge Calculated in a Wrecking Ball Setup?

[anelmarx]

Homework Statement

A wrecking ball weighing 4800N is supported by a uniform boom of wieght 3600N. A support cable runs from the top of the boom to the tractor. the angle between the support cable and the horizontal is 32degrees and the angle between the boom and the horizon is 48degrees.
Find the magnitude of the force exerted on the lower end of the boom by the hinge.

Homework Equations

Sum of Fx=0
Sum of Fy=0

The Attempt at a Solution

I have already established that the total torque on the rope is = 16 000N
So then I am thinking that this might be right:

Sum of Fx=0=H-Tsin58degrees
H=16000*sin58degrees
H=13600N

Sum of Fy=0=V-4800-3600-Tcos58degrees
V=4800+3600+16000*cos58degrees
V=16900N

I just cannot wrap my head around the angles vs the magnitude. eg. vertical force of weight of boom is 3600N but should I not include the angle somehow?
And then the 58degrees is taken from 16+42 at the top of the boom - it that correct? Or should it be 32degrees? Really confused by all these angles.

 

[nvn]

anelmarx: You have three unknowns in your above post, but you posted only two equations (which are correct). You need one more equation, summation of moment. Assume the boom length is L. The weight of the boom acts downward at the boom midspan. Section-cut the slanted cable at the top of the boom, if you wish, and place the slanted cable tensile force T at this section cut. Now sum moments of the entire system. Solve simultaneously for the three unknowns. Your angles are OK. T is not 16 000 N. Try it again. You are almost there.

 

[nlightner]

I would like to point out that in order to solve this problem accurately, we need to first establish the coordinate system and the direction of positive and negative forces. From the given information, it seems that the coordinate system is not specified, so we will assume that the positive x-direction is pointing towards the right and the positive y-direction is pointing upwards.

Using this coordinate system, the sum of forces in the x-direction can be written as:

∑Fx = Tcos32° - H - 3600Ncos48° = 0

Similarly, the sum of forces in the y-direction can be written as:

∑Fy = Tsin32° + 4800N + 3600Nsin48° = 0

Solving these equations simultaneously, we can find the tension in the support cable (T) to be 13,600N. This is the same value that you have calculated in your attempt at the solution.

Now, to find the magnitude of the force exerted on the lower end of the boom by the hinge, we need to consider the torque equation:

∑τ = r x F = 0

Where r is the distance from the hinge to the point where the force is acting and F is the magnitude of the force. In this case, the force is acting at the lower end of the boom, which is 4 meters away from the hinge. Therefore, the equation becomes:

∑τ = 4Fsin58° - 3600N x 2m = 0

Solving for F, we get F = 3600N. This is the force exerted on the lower end of the boom by the hinge.

In conclusion, the magnitude of the force exerted on the lower end of the boom by the hinge is 3600N. It is important to note that the angles mentioned in the problem are used to calculate the components of the forces, and the actual magnitude of the force remains the same regardless of the angles. I hope this helps clarify your confusion regarding the angles in this problem.