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(B) To keep the block at rest on the smooth inclined plane,the forces acting on the block must be balanced.Along the inclined plane,the component of t

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$\frac{3 mg}{4}$

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(B) To keep the block at rest on the smooth inclined plane,the forces acting on the block must be balanced.Along the inclined plane,the component of the gravitational force acting downwards is $mg \sin 37^{\circ}$.The horizontal force $F$ is applied,and its component along the inclined plane acting upwards is $F \cos 37^{\circ}$.For the block to be in equilibrium,these two forces must be equal:$F \cos 37^{\circ} = mg \sin 37^{\circ}$Using the values $\cos 37^{\circ} = \frac{4}{5}$ and $\sin 37^{\circ} = \frac{3}{5}$:$F \left( \frac{4}{5} \right) = mg \left( \frac{3}{5} \right)$$F = \frac{3 mg}{4}$

$A$ block of mass $m$ is placed on a smooth fixed inclined plane with an angle of inclination of $37^{\circ}$. $A$ horizontal force $F$ is applied to keep the block at rest. Find the value of $F$.

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