A block of mass M is on a rough horizontal surface with coefficient of friction μ = 3. A horizontal force F equal to 4/5 of the minimum force required to just move the block is applied. Find the magnitude of the force exerted by the ground on the block.

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G1 Physics · Accepted Answer

Minimum force to start motion is μMg = 3Mg, so the applied force is F = (4/5)(3Mg) = 2.4Mg. Since the block is still static, friction equals 2.4Mg and normal reaction is Mg. Ground contact force = √[(Mg)² + (2.4Mg)²] = 2.6Mg.