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

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(B) From the free body diagram, the forces acting on the body are resolved into horizontal and vertical components.For the body to be on the verge of

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$20 \,N, 30 \,N$

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(B) From the free body diagram, the forces acting on the body are resolved into horizontal and vertical components.For the body to be on the verge of moving, the horizontal component of the applied force must balance the frictional force $(f)$:$f = F \cos 45^{\circ}$$f = 28.28 	imes \frac{1}{\sqrt{2}} = 28.28 	imes 0.707 = 20 \,N$The vertical forces must be in equilibrium, so the normal reaction $(R)$ plus the vertical component of the applied force must equal the weight of the body $(W = 50 \,N)$:$R + F \sin 45^{\circ} = 50$$R = 50 - 28.28 \sin 45^{\circ}$$R = 50 - 28.28 	imes \frac{1}{\sqrt{2}} = 50 - 20 = 30 \,N$Thus, the frictional force is $20 \,N$ and the normal reaction is $30 \,N$.

$A$ body of weight $50 \,N$ is placed on a horizontal surface as shown in the figure. The minimum force required to move the body is $28.28 \,N$. The frictional force and the normal reaction are respectively.

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