- #1

sinequanon

- 6

- 0

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?