- #1

adam199

- 17

- 0

Picture of the problem:

Relevant equations are the ones used to find the reactions at the supports (moment about points, forces in the x and y) and the ones used to determine the centroid.

I'm having a hard time finding a. I found the moment of area for each triangle, with the origin at A, and the sum of the areas turned out to be ƩA=1200+600a and ƩxA=200a^2-400a+3200. I know that the coordinate X of the centroid can be found by dividing ƩxA by ƩA, and doing so brought me no closer to finding a. I then tried to set up the equations for the moment about point A, Fy, and Fx and got:

Ma=500a^2-400a+3200-4By=0 (By is the reaction force at point B in the y-direction)

Fx=0, Ax=0 (since all other forces, including the roller at point B, are vertical)

Fy=Ay+By-(1200+600a)=0 (where the 1200+600a is the combined load applied to the beam, figured out from the centroid stuff earlier)

I can't figure out how to manipulate the centroid or moment/force equations to obtain a. Any help would be much appreciated.

Relevant equations are the ones used to find the reactions at the supports (moment about points, forces in the x and y) and the ones used to determine the centroid.

I'm having a hard time finding a. I found the moment of area for each triangle, with the origin at A, and the sum of the areas turned out to be ƩA=1200+600a and ƩxA=200a^2-400a+3200. I know that the coordinate X of the centroid can be found by dividing ƩxA by ƩA, and doing so brought me no closer to finding a. I then tried to set up the equations for the moment about point A, Fy, and Fx and got:

Ma=500a^2-400a+3200-4By=0 (By is the reaction force at point B in the y-direction)

Fx=0, Ax=0 (since all other forces, including the roller at point B, are vertical)

Fy=Ay+By-(1200+600a)=0 (where the 1200+600a is the combined load applied to the beam, figured out from the centroid stuff earlier)

I can't figure out how to manipulate the centroid or moment/force equations to obtain a. Any help would be much appreciated.

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