Question

Three concurrent coplanar forces 1N, 2N and 3N acting along different directions on a body:
A. Can keep the body in equilibrium if 2N and 3N act at right angles.
B. Can keep the body in equilibrium if 1N and 2N act at right angles.
C. Cannot keep the body in equilibrium.
D. Can keep the body in equilibrium if 1N and 3N act at an acute angle.

Answer 1

Hint: Use the resultant of two vectors formula $R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$, where R is the magnitude of the resultant vector, A and B are the magnitudes of the given two vectors acting on the body and $\theta$ is the angle between vectors A and B.

Complete step-by-step answer:
The question is asked how the three concurrent coplanar forces with magnitudes 1N, 2N and 3N acting along different directions can keep a body in equilibrium. To solve these types of questions there are two ways. One is to directly find the answer. Another is that we can go through the given options one by one and check which one gives the required result. The second method looks easier so let us go with that one.
Check the first option. It is given that the body will be in equilibrium if 2N and 3N act at right angles. Suppose the 2N and 3N forces act at a right angle and the force of 1N acts at any random point. The resultant force will be $\sqrt{{{2}^{2}}+{{3}^{2}}+2(2)(3)\cos (90)}=\sqrt{4+9+0}=\sqrt{13}N$.
So, now there are two forces of 1N and $\sqrt{13}\text{N}$. The only way for two forces to balance each other is that they must be of equal magnitude and opposite in direction to each other. Therefore, in this situation the body cannot be in equilibrium.
Let us check the second option. Follow the same procedure as done for option one. Here the resultant force will be $\sqrt{{{1}^{2}}+{{2}^{2}}+2(1)(2)\cos (90)}=\sqrt{1+4+0}=\sqrt{\text{5}}\text{N}$. Now the two forces remaining are 3N and $\sqrt{\text{5}}\text{N}$. $3\ne \sqrt{5}$, therefore, the body will not be in equilibrium.
Let us keep option C for later. Option D tells that the body can be in equilibrium if the forces of 1N and 3N act at an acute angle. Suppose 1N and 3N act at an angle $\theta$. The resultant of this two will be $\sqrt{{{1}^{2}}+{{3}^{2}}+2.(1).(3)\cos \theta }=\sqrt{10+6\cos \theta }$. Now, if this resultant force balances the third force of 2N then, $2=\sqrt{10+6\cos \theta }$
$\Rightarrow 4=10+6\cos \theta$. Solve for $\theta$.
$-6=6\cos \theta \Rightarrow \cos \theta =-1$. Therefore, $\theta =\pi$ i.e. $180^o$. However, $\pi$ is not an acute angle. Therefore, this option is also incorrect. Hence, the correct option is (C) cannot keep the body in equilibrium.

Note: One may think that even option C is also incorrect since the body is in equilibrium when angle between the forces of 1N and 3N acting on the body is 180 degrees, as you can see where we solved for the third option. It is true that the body will be in equilibrium if the angle between the forces of 1N and 3N is 180 degrees and if the 2N force acts along the direction of 1N. However, it is told in the question that all the three forces are acting along different directions. Therefore, the above situation is not possible.