Question

Three forces are acting on a particle of
mass m initially in equilibrium. If the first
2 forces (R, and R) are perpendicular to
each other and suddenly the third force
(R3) is removed, then the acceleration of
the particle is:​

Answer 1

Let the forces,be P,Q,R,and the resultant of P and Q be PQ

Now,to maintain the equilibrium,,the resultant of P and Q,should be equal and opposite,to the third force(as shown in figure)

Resultant of P and Q=(a^2+b^2+2abcos∅)^1/2

PQ=(a^2+b^2+2abcos90)^1/2......i)

we , know,cos90*=0

putting this value in PQ,we get

PQ=(a^2+b^2)^1/2

As, mention above,third force=-(a^2+b^2)^1/2....(- sign is used to show, opposite direction)

Now,If third force R is removed,than

Resultant force=PQ=(a^2+b^2)^1/2

mass of the body=m

force=Mass× ACCELERATION

a=m/f

a=m/(a^2+b^2)^1/2

Hence the acceleration of the particle will be,m/(a^2+b^2)^1/2

{hope it helps}