A body of weight 50 ,N placed on a horizontal | vedclass

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(B) To find the frictional force and normal reaction,we resolve the applied force $F = 28.2 \,N$ into its horizontal and vertical components.Horizonta

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$20 \,N, 30 \,N$

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(B) To find the frictional force and normal reaction,we resolve the applied force $F = 28.2 \,N$ into its horizontal and vertical components.Horizontal component: $F_x = F \cos 45^\circ = 28.2 	imes \frac{1}{\sqrt{2}} = 28.2 	imes 0.707 \approx 20 \,N$.Since the body is just moved,the frictional force $f$ is equal to the horizontal component of the applied force: $f = 20 \,N$.Vertical component: $F_y = F \sin 45^\circ = 28.2 	imes \frac{1}{\sqrt{2}} = 28.2 	imes 0.707 \approx 20 \,N$.The normal reaction $R$ is given by the balance of vertical forces: $R + F_y = W$,where $W = 50 \,N$ is the weight of the body.$R = W - F_y = 50 \,N - 20 \,N = 30 \,N$.Thus,the frictional force is $20 \,N$ and the normal reaction is $30 \,N$.

$A$ body of weight $50 \,N$ placed on a horizontal surface is just moved by a force of $28.2 \,N$ applied at an angle of $45^\circ$ with the horizontal. The frictional force and the normal reaction are

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