# Homework Help: Conditions of equilibrium problem

1. Jul 12, 2010

### Dragonite

1. The problem statement, all variables and given/known data
Consider the following cantilevered beam:
The beam has a mass of m = 25 kg and is 2.2 meters long. The suspended block has
a mass M = 280 kg and the supporting cable makes an angle of 30 degrees with the beam.
Determine the force that the wall exerts on the beam at the hinge and determine
the tension in the supporting cable.

(There's a horizontal beam making a 90 degree angle with a wall. At the end of the beam, there's a string with the 280 kg block. Also, there's a string connecting the top of the wall to the end of the beam. )

2. Relevant equations

∑F=0
∑T=0

3. The attempt at a solution

1) I converted the weight of the block and the beam to newtons.

280 kg x 9.8 m/s^2 = 2744 N
25 kg x 9.8 m/s^2 = 245 N

2) Using the first condition of equilibrium,

245 + 2744 = 2989 N = Downward force = Upward force by string making 30 degree angle with beam

3) Using sine law

Force toward the left of the string = 2989 sin 60/sin 30 = 5177.1 N = Force toward the right by the wall

4)

Squareroot of quantity Fx^2 + Fy^2 = 5978 N = Tension on string

2. Jul 12, 2010

### kuruman

So far so good.

What about the vertical force exerted by the hinge on the beam? There is also a horizontal force exerted by the hinge on the beam.

Bottom line: You have three unknowns, the tension, the vertical component of the hinge force and the horizontal component of the hinge force. Therefore, you need to write down three equations to solve for the unknowns. What are the three equations?

3. Jul 19, 2010

### Dragonite

Let me try...

Horizontal force of hinge = Horizontal component of tension / Tension sin 60
Vertical component of tension + Vertical force of hinge = 2989 N
2744 N x 2.2 + 245 x 1.1 = Vertical component of tension / Tension sin 30

Is that right?

4. Jul 19, 2010

### kuruman

Where did you get this? Where does it say that the sum of the horizontal forces is zero? Your equation must be of the form
F1x+F2x+F3x+... = 0.
Correct. This is the same as F1y+Ty+(-W1)+(-W2)=0.
Note that it is in the correct form, ΣFy=0.
What is the torque generated by the tension? It is not what you have on the right side.

5. Jul 19, 2010

### Dragonite

Wouldn't the horizontal forces be equal since the tension and the beam are in equilibrium?

Ok.

Oh sorry. I forgot to put the distance.

2744N x 2.2 + 245N x 1.1 - Ty x 2.2 = 0

6. Jul 19, 2010

### kuruman

Yes they would. Does your equation

Horizontal force of hinge = Horizontal component of tension / Tension sin 60

say this or does it say something else? Why are you dividing by Tension sin60?

7. Jul 21, 2010

### Dragonite

I meant to say horizontal component of tension or tension sin 60. Sorry and thank you very much.