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A bar of 3m. Length & negligible weight rests in horizontal position on two smooth inclined. (fig.)

Determine the distance at which the lead $Q = 150N$ should be placed from point B to keep the bar horizontal

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1 Answer
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The bar is supported by two plane surfaces. Plane surfaces can give only normal reaction (reaction perpendicular to the surface.)

Hence, the FBD can be drawn as follows

enter image description here

Given: the bar should be horizontal. It means that net moment of the system should be zero. This condition will help us …

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