A force vec{F} = hat{i} + 4hat{j} text{ N} ac | vedclass

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(A) $1$. Analyze the forces acting on the block in the vertical $(y)$ direction. The forces are the gravitational force ($mg = 1 	ext{ kg} 	imes 10

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$-\hat{i} 	ext{ N}$

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(A) $1$. Analyze the forces acting on the block in the vertical $(y)$ direction. The forces are the gravitational force ($mg = 1 	ext{ kg} 	imes 10 	ext{ m/s}^2 = 10 	ext{ N}$ downwards),the vertical component of the applied force ($F_y = 4 	ext{ N}$ upwards),and the normal force ($N$ upwards).$2$. Equilibrium in the vertical direction: $N + F_y = mg \implies N + 4 = 10 \implies N = 6 	ext{ N}$.$3$. Calculate the maximum limiting friction: $f_{max} = \mu N = 0.3 	imes 6 = 1.8 	ext{ N}$.$4$. Analyze the forces in the horizontal $(x)$ direction. The applied horizontal force is $F_x = 1 	ext{ N}$.$5$. Since the applied horizontal force $(1 	ext{ N})$ is less than the maximum limiting friction $(1.8 	ext{ N})$,the block remains at rest (static equilibrium).$6$. In static equilibrium,the frictional force exactly balances the applied horizontal force: $f = -F_x = -1 \hat{i} 	ext{ N}$.

$A$ force $\vec{F} = \hat{i} + 4\hat{j} \text{ N}$ acts on the $1 \text{ kg}$ block shown in the figure. The coefficient of friction between the block and the surface is $\mu = 0.3$. The force of friction acting on the block is:

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