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13 March, 13:26

Find the minimum weight of block A in order for motion to be impending down the plane (to the left). Assume the pulley to be frictionless. Given: WB = 275 lbs, θ = 35 °, μ = 0.25 for all surfaces.

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  1. 13 March, 13:47
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    88.25lbs

    Explanation:

    Let N be the minimum weight to be found.

    firstly we have to solve for the force body diagram for both blocks, for block B which is stable, we have its weight acting downwards, for it to move left affected by block A it would need a horizontal component which is the frictional force and we can calculate it as weight of B x coefficient of friction = 275lbs x 0.25 = 68.75lbs.

    Calculating same for block A: (free body diagram)

    the horizontal and vertical component of the forces on A would be:

    Nsin35 (horizontal comp) + Ncos35 (0.25) (vertical comp) = 68.75lbs

    =0.574N+0.205N=68.75lbs.

    solving the above equation and making N subject of formula would result to 88.25lbs.
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