What is the force acting on the fixed points of an inclined bar?

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SUMMARY

The force acting on the fixed points of an inclined bar secured between two walls is equal to the component of the bar's weight acting perpendicular to the walls. The equilibrium condition, represented by the equation ΣF=0, indicates that the forces at the fixture points are equal and opposite to maintain stability. The load on each fixture point is parallel to the weight of the bar, which acts at its midpoint. This analysis confirms that the forces exerted on the walls are a direct result of the bar's angle and weight distribution.

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Homework Statement


We have a straight bar of specific length that its edges are secured between two walls. However the bar is not parallel to the ground, but has an angle. My question is, what is the force that pushes each of the two fixed points in the walls, down?

Homework Equations


ΣF=0

The Attempt at a Solution


I think that the load to the fixture points will be parallel to the weight of the bar acting at its middle point. And the forces will be equal. Is that true?
 
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