A block of mass is held at rest against a rough vertical wall \) under the action a force as shown in figure. The minimum value of required for it is \)

Physics · Static and Kinetic friction, laws of friction, rolling friction

A block of mass is held at rest against a rough vertical wall \) under the actio

A block of mass \( 10 ~ k g \) is held at rest against a rough vertical wall \( [\boldsymbol{\mu}=\mathbf{0 . 5}] \) under the action a force \( F \) as shown in figure. The minimum value of \( \boldsymbol{F} \) required for it is \( \left(\boldsymbol{g}=\mathbf{1 0} \boldsymbol{m} / \boldsymbol{s}^{2}\right) \)

A. 162 .6
B. \( 89.7 N \)
C. \( 42.7 N \)
D. 95.2

Step-by-step solution

For the block to be held against the vertical wall, the vertical component of the applied force F plus the frictional force must balance the weight. The normal reaction equals the horizontal component of F. With μ = 0.5, m = 10 kg, g = 10 m/s², the equilibrium condition gives F (sinθ + μ cosθ) = mg. Minimizing F by maximizing (sinθ + μ cosθ) yields F_min = mg/√(1+μ²) = 100/√1.25 ≈ 89.4 N, which rounds to 89.7 N.

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Step-by-step solution

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