# Engineering Statics Question Regarding Finding Normal Force

1. Jun 3, 2014

### Temp0

1. The problem statement, all variables and given/known data
http://i.imgur.com/rwbXqW8.png

2. Relevant equations

3. The attempt at a solution
So I started off by analyzing the pulley, and determined that each cable carries 100N of force. After that, I drew the FBD of the beam, where there is a 100N force going downwards on the right end and the normal force perpendicular to the beam on the left end. I took the moment about point B, and since this is in equilibrium, the moment should be zero. Therefore I came up with the equation:

-100cos(30)(0.3) + Nwall (0.1) = 0
Where Nwall is the normal force provided by the contact with the wall.
I came up with the answer 259.8N, can anyone check my work please? Thank you in advance.

2. Jun 3, 2014

### PhanthomJay

If the normal force at the left end was perpendicular to the beam, then there would be a component of that force acting parallel to the wall. But the wall is frictionless. So........?

3. Jun 3, 2014

### Temp0

Yeah I noticed my mistake, the beam will create a contact force that's perpendicular to the wall and I need to figure out the angle for the component perpendicular to the beam where I can then take the moment.

4. Jun 3, 2014

### PhanthomJay

Yes, that will work.

5. Jun 3, 2014

### Temp0

So, for my final equation, I found that the normal force of the wall is 30 degrees from the perpendicular component to the beam, so I have:
The sum of moments about point B = Nwall cos30 (0.1) - 100cos30(0.3) = 0
Getting an Nwall of 300.

6. Jun 3, 2014

### PhanthomJay

Looks good!
I suppose you should indicate its sense....whether the force of the beam on the wall perpendicular to the wall is acting up and to the left or down and to the right... ?