# Beam and Hinge Confusion

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. ΣFx = Rx - TcosΘ = 0

2. ΣFy = Ry + TsinΘ - Fobject - Fbeam = 0

3. TsinΘ(dcable) - Fbeam(dbeam) - Fobject(dobject)

*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.

ΣFx = Rx - 3116.68cos60 = 0
Rx = 2968.36 N

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

ΣFy = Ry + 3116.68sin60 - 1000 - 500 = 0
Ry = 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?

Related Introductory Physics Homework Help News on Phys.org
rl.bhat
Homework Helper
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.

Last edited:
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?

Rx - 1828.28cos60=0
Rx=914.14 N

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