Question

A body of mass 2kg is placed on a smooth horizontal surface. Two forces F1 = 20N , F2 = 10√3N 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

Answer 1

Answer:

20N

25N

5N

zero

Answer :

D

Solution :

Let us find vertical force:

F1sin30∘+F2sin60∘

=20×12+103–√(3–√2)=25N

This is greater than mg. So contact of body with surface will be lost and reaction will be zero.

Answer 2

Given:

The mass of the body = 2kg

The magnitude of force F1 = 20N

The magnitude of force F2 = 10√3N

To Find:

The reaction of the surface on the body

Solution:

The normal reaction of the surface as experienced by the body is equal to the vertical force applied on the surface.

The vertical forces applied on the surface are:

1. Weight of the body = mg = 2 X 10 = 20N
2. The vertical component of F1 = F1 sin 30° =  $20 X\frac{1}{2}$

       = 10N

    3.The vertical component of F2 = F2 sin 60° = $10\sqrt{3} X \frac{\sqrt{3} }{2}$

      = 15N

So the net vertical force = 20 + 10 + 15 = 45N

Hence, the reaction of the surface on the body will be 45N.