First slide

Motion of block on Smooth incline

Question

A body of mass 2.0 kg is placed on a smooth horizontal surface. Two forces $F_{1}$ = 20 N and $F_{2} = 10 \sqrt{3}$ N are acting on the body in directions making angles of 30° and 60° to the surface. The reaction of the surface on the body will be

Moderate

A

20 N
B

25 N
C

5 N
D

zero

Solution

Let us find vertical force:
$F_{1} \sin 30^{0} + F_{2} \sin 60^{0}$ = $20 \times \frac{1}{2} + 10 \sqrt{3} (\frac{\sqrt{3}}{2}) = 25 N$
This is greater than mg. So contact of body with surface will be lost and reaction will be zero.

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