What Is the Reacting Force From the Hinge and Its Angle?

[sinequanon]

Homework Statement

A 15 meter beam jutting out of the side of a building is held by a hinge (at the wall) and a cable at 10 meters from the wall. The angle between the beam and the cable is 60 degrees and the mass of the beam is 250 kg. If a 1000 N object is located on the beam, 7 meters from the wall, what is the reacting force R from the hinge and at what angle is it applied? Assume the system is in equilibrium.

Homework Equations

1. ΣF_x = R_x - TcosΘ = 0

2. ΣF_y = R_y + TsinΘ - F_object - F_beam = 0

3. TsinΘ(d_cable) - F_beam(d_beam) - F_object(d_object)

*Use 10 m/s² for the value of gravitational acceleration.

The Attempt at a Solution

Alright, so I just wanted to double check to see if I'm actually doing this correctly.

First I substitute into the third equation in order to find the cable tension.

Tsin60(10 m) - (2500 N)(7.5 m) - (1000 N)(7 m) = 0
T = 2973.44

Then, I would substitute the T value into the other equations.

ΣF_x = R_x - 2973.44cos60 = 0
R_x = 1486.72 N

ΣF_y = R_y + 2973.44sin60 - 1000 - 2500 = 0
R_y = 924.925 N

From here, it appears to be a simple matter of using the Pythagorean Theorem and then just using inverse cosine to find Θ_R.
R = $$\sqrt{1486.72^2 + 924.925^2}$$ = 1750.95 N

cos^-1Θ = 1486.72/1750.95
Θ = 31.88°

I was hoping someone would be able to double check to see if my understanding of this matter is correct or otherwise. I was also wondering if someone could tell if my final answer R should be positive or negative, as that is one thing I haven't a clue about.

 

[PhanthomJay]

Looks real good to me. Often it is best to leave the reaction force at the hinge in terms of its x and y components, that is, Rx = +1486 N (or 1486 N pointing right) , and Ry = +925 N (or 925 N pointing up). The sign of the reaction force components is largely a matter of convention; its direction as shown on a sketch is the important part. Now since the problem asked you to provide the Resultant force and angle at the hinge, your calc is correct for the magnitude of that value: R = 1751 N pointing up and to the right at a 32 degree angle with the horizontal axis. The resultant force really is a magnitude only without a sign, the direction shown is is what is important. Good work!