Force of Bar on wall

1. The problem statement, all variables and given/known data
How hard does the beam push inward on the wall?

2. Relevant equations

3. The attempt at a solution

I know that the tension on the cable is 1.32*10^4 N

From there I turned the weight and tension into x/y components and tried to add the x components, so that:
1.32*10^4cos(40) + 14700sin(30) = 17461.79 N

I don't understand why this is not correct?
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 Recognitions: Gold Member Homework Help Science Advisor what is the weight and length of the bar?
 Sorry, the bar is 8m long, and 1500kg.

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Force of Bar on wall

There are no z components, only x an y components. The weight has a y component only. The tension force has x and y components. The force of the wall on the bar has x and y components. Use Newton 1 in the x and y directions to solve for the x and y components of the wall force on the bar. Your tension force is correct.
 Thanks for taking the time to respond Jay, That z component was a typo, I meant to say x-component. I believe I did what you described above. I got the reaction force as shown in this image: (increasing x is down an left) What exactly am I doing incorrectly?
 Recognitions: Gold Member Homework Help Science Advisor The reaction force at the wall is not directed parallel to the bar...that only occurs for 2 force members , like the cable. You'll have to include the componets perpendicular to the bar also. You also do not have to break up the components into forces parallel and perpendicular to the incline, although that is acceptable. But you can also look at just x and y (horiz and vert) components.
 Recognitions: Homework Help The bar presses on the wall at an angle (30° to the horizontal). Some of the force is directed downward (pushing the wall down into the ground), some is directed "inward".
 Like this? This solution of 13000 N is also incorrect. I'm not sure what you mean by breaking it up into components perpendicular to the bar. I didn't think that I could use torque to find the force for this question. Alternatively, 17461.8cos(30), my original reaction force that was angled to the wall, resolved into it's x component, produces 15122.36?

Recognitions:
Gold Member
Homework Help
 Okay, so: $$mg-R_{y}-Tsin(80)=0$$ $$R_{y} =mg-Tsin(80)$$ $$R_{y} =1500(9.8)-(1.32*10^{4})sin(80)$$ $$R_{y} = 1700.54$$ So now my Y component is: 1700 N, and my x is 13000 N. Do they want the sum of the two for this "inward" force from the bar on the wall? It's not simply the x component as I initially thought. Thanks.