Question

Three forces are acting on a particle of mass ‘m’ initially in equilibrium. If the first two forces$\left( {{{\text{R}}_1}{\text{ and }}{{\text{R}}_2}} \right)$are perpendicular to each other and suddenly the third force$\left( {{{\text{R}}_3}} \right)$is removed, then the acceleration of the particle is?
$
  {\text{A}}{\text{. }}\dfrac{{{{\text{R}}_3}}}{{\text{m}}} \\
  {\text{B}}{\text{. }}\dfrac{{{{\text{R}}_1} + {{\text{R}}_2}}}{{\text{m}}} \\
  {\text{C}}{\text{. }}\dfrac{{{{\text{R}}_1}{\text{ - }}{{\text{R}}_2}}}{{\text{m}}} \\
  {\text{D}}{\text{. }}\dfrac{{{{\text{R}}_1}}}{{\text{m}}} \\
$

Answer 1

Hint: In order to find the acceleration of the particle, we use the concept of equilibrium. That is, the sum of all the forces or the net force acting on a body at equilibrium is zero.

Complete Step-by-Step solution:
Given Data,
First two forces are perpendicular and the third force is removed.
The given forces in vector form are$\mathop {{{\text{R}}_1}}\limits^ \to ,\mathop {{{\text{R}}_2}}\limits^ \to {\text{ and }}\mathop {{{\text{R}}_3}}\limits^ \to $.
And $\mathop {{{\text{R}}_1}}\limits^ \to {\text{ and }}\mathop {{{\text{R}}_2}}\limits^ \to $are perpendicular to each other.

We know the magnitude of a vector $\mathop {\text{A}}\limits^ \to $is given as |A| (modulus A) and it denotes just the scalar quantity of the vector.
Hence the magnitudes of the given forces are $|{{\text{R}}_1}|,{\text{ |}}{{\text{R}}_2}|{\text{ and |}}{{\text{R}}_3}|$respectively.

Initially the body is in equilibrium. Equilibrium is a state in which all the forces influencing the body are balanced or cancel each other out. There will be no change at equilibrium.
Hence the sum of all the forces acting on the body is equal to zero.
\[
   \Rightarrow \mathop {{{\text{R}}_1}}\limits^ \to + \mathop {{{\text{R}}_2}}\limits^ \to {\text{ + }}\mathop {{{\text{R}}_3}}\limits^ \to = 0 \\
   \Rightarrow \mathop {{{\text{R}}_1}}\limits^ \to + \mathop {{{\text{R}}_2}}\limits^ \to = - \mathop {{{\text{R}}_3}}\limits^ \to \\
\]
Therefore in terms of magnitude this is –
$|{{\text{R}}_1} + {{\text{R}}_2}|{\text{ = |}}{{\text{R}}_3}|$ ------- (1)
The negative sign does not matter because anything put in the modulus function only gives us a positive value and magnitude just means the scalar quantity not the sign.

Now if the force \[\mathop {{{\text{R}}_3}}\limits^ \to \] is removed the body is subject to the remaining forces \[\mathop {{{\text{R}}_1}}\limits^ \to \] and \[\mathop {{{\text{R}}_2}}\limits^ \to \], the net-force acting on the body is given by:
\[\mathop {{{\text{F}}_{{\text{net}}}}}\limits^ \to = \mathop {{{\text{R}}_1}}\limits^ \to + \mathop {{{\text{R}}_2}}\limits^ \to \] ----- (2)
So this force acted upon a body of mass ‘m’, causing some acceleration on the body. Acceleration of a body of mass subject to a force is given by:
\[\mathop {\text{a}}\limits^ \to = \dfrac{{\mathop {\text{F}}\limits^ \to }}{{\text{m}}}\]
It is also a vector quantity.
The magnitude of acceleration is\[|\mathop {\text{a}}\limits^ \to | = \dfrac{{|\mathop {\text{F}}\limits^ \to |}}{{\text{m}}}\].
In our case the mass of the body is ‘m’ and the force acting on it is the net-force. (Equation -2)
\[|\mathop {\text{a}}\limits^ \to | = \dfrac{{|\mathop {{{\text{R}}_1}}\limits^ \to + \mathop {{{\text{R}}_2}}\limits^ \to |}}{{\text{m}}}\]
Substituting equation (1) in the above we get –
\[|\mathop {\text{a}}\limits^ \to | = \dfrac{{|\mathop {{{\text{R}}_3}}\limits^ \to |}}{{\text{m}}}\]
Therefore the acceleration on the particle is$\dfrac{{{{\text{R}}_3}}}{{\text{m}}}$. Option A is the correct answer.

Note – In order to answer this type of questions the key is to understand the concepts of force and equilibrium and the condition of an equilibrium. Force is defined as the energy that causes motion on an object, it is a physical attribute of movement. Its S.I unit is “Newton”. The formula of force acting on a body of mass ‘m’ to produce an acceleration ‘a’ in it is given by “F = ma”. When all the forces acting on a body cancel each other, the body stays still, it is said to be in equilibrium.