A force vec{F}=hat{i}+4 hat{j} , N acts on th | vedclass

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(A) $1$. First,calculate the normal force $N$ acting on the block. The vertical forces are the normal force $N$ (upward),the weight $mg$ (downward),an

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$-\hat{i} \, N$

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(A) $1$. First,calculate the normal force $N$ acting on the block. The vertical forces are the normal force $N$ (upward),the weight $mg$ (downward),and the $y$-component of the applied force $F_y = 4 \, N$ (upward).$2$. Equilibrium in the vertical direction: $N + F_y = mg \implies N + 4 = (1)(10) \implies N = 6 \, N$.$3$. Calculate the limiting friction $f_L = \mu N = 0.3 	imes 6 = 1.8 \, N$.$4$. The horizontal component of the applied force is $F_x = 1 \, N$.$5$. Since the applied horizontal force $F_x = 1 \, N$ is less than the limiting friction $f_L = 1.8 \, N$,the block remains at rest.$6$. For a block at rest,the static friction force exactly balances the applied horizontal force. Therefore,the friction force $f = -F_x = -1 \hat{i} \, N$.

$A$ force $\vec{F}=\hat{i}+4 \hat{j} \, N$ acts on the block of mass $1 \, kg$ 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 (Take $g = 10 \, m/s^2$)

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