A block of mass M is placed on a rough surfac | vedclass

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(A) The minimum force required to just move the block is the limiting friction,$f_L = \mu N = \mu Mg = 3Mg$.Given that the applied force $F = \frac{4}

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$2.6\ Mg$

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(A) The minimum force required to just move the block is the limiting friction,$f_L = \mu N = \mu Mg = 3Mg$.Given that the applied force $F = \frac{4}{5} f_L = \frac{4}{5} (3Mg) = 2.4Mg$.The normal force exerted by the ground is $N = Mg$.The frictional force acting on the block is $f = F = 2.4Mg$ (since $F The total force exerted by the ground on the block is the resultant of the normal force and the frictional force:$R = \sqrt{N^2 + f^2} = \sqrt{(Mg)^2 + (2.4Mg)^2} = \sqrt{Mg^2 + 5.76Mg^2} = \sqrt{6.76Mg^2} = 2.6Mg$.

$A$ block of mass $M$ is placed on a rough surface with a coefficient of friction $\mu = 3$. If the applied force $F$ is $(4/5)$ of the minimum force required to just move the block,find the total force exerted by the ground on the block.

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