Engineering Statics Question Regarding Finding Normal Force

[Temp0]

Homework Statement

http://i.imgur.com/rwbXqW8.png

Homework Equations

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.

 

[PhanthomJay]

Homework Statement

http://i.imgur.com/rwbXqW8.png

Homework Equations

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.

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...?

 

[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.

 

[PhanthomJay]

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.

Yes, that will work.

 

[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.

 

[PhanthomJay]

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 N .

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... ?