Beam and Hinge Confusion

[sinequanon]

1. Homework Statement

A 10 meter beam with a mass of 100 kg is hinged at the wall with a supporting cable at 6 meters from the hinge, making a 60 degree with the beam. There is a 50 kg person located at 9 meters from the hinge. Assuming equilibrium, what is the tension on the cable, the force on the hinge, and the angle of the reactant force.

2. Homework Equations

1. Σ_Fx = R_x - TcosΘ = 0

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

3. TsinΘ(d_cable) - Fbeam(d_beam) - Fobject(d_object)

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

3. 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) - (1000 N)(5 m) - (500 N)(9 m) = 0
T = -3116.68 N

My first question is with that bolded segment: is it correct to have it subtracted if I'm looking for T? Should it be added instead since the direction is clockwise?

If that's right, then I need to find the force on the hinge, and here is where I really find trouble.

Σ_Fx = R_x - 3116.68cos60 = 0
R_x = 2968.36 N

Okay, that part was easy for me, but...

Σ_Fy = R_y + 3116.68sin60 - 1000 - 500 = 0
R_y = 1199.12

I asked my instructor about this, but he insists this is incorrect (which I'm sure it is). He is telling me to use

6F +1000(1) = 500(3)

I have only a vague understanding of what is involved in this equation. Could someone help me wrap my head around this a little better?

 

[rl.bhat]

Tsin60(10 m) - (1000 N)(5 m) - (500 N)(9 m) = 0

The cable is attached at 6m from the hinges. So check the equation again.

6F +1000(1) = 500(3)

With this equation you can find the force on hinges. Here the moments of couple are taken about the point of suspension of the cable to the beam.

 

[sinequanon]

Ah. Well, that was pretty careless of me. Thanks for that.

Okay, so...

Tsin60(6 m) - (1000 N)(5 m) - (500 N)(9 m) = 0
T=1828.28

I take it I would be correct with using subtraction in that equation then?

R_x - 1828.28cos60=0
R_x=914.14 N

However, I still would like confirmation on which equation to use if finding the force acting on the hinge.


Sorry, just saw your edit.

Ah. That explains it so much better. So that will solve for the force on the hinges but will I still need to find R_y in order to find the angle of the reactant force?

 

[rl.bhat]

Find F. Using F and Rx you can find angle of reactant force.