Question

When a force $F$ acts on a body of mass $m$, then the acceleration produced in the body is $a$. If three equal coplanar forces $F_1=F_2=F_3=F$ act on the same body as shown in the figure, the acceleration produced is:

• A. $(\sqrt{2}-1)a$
• B. $(\sqrt{2}+1)a$
• C. $\sqrt{2}a$
• D. $a$

Answer 1

The correct option is A $(\sqrt{2}-1)a$

From the given figure,
The resultant of the forces $F_1$ and $F_2$ will be $=\sqrt{2}F$.
Now,

Since $F_3$ is ${135}^o$ to $F_2$ and resultant of $F_{1}and{F}_2$ is ${45}^o$ to $F_1$

i.e. both are in exactly opposite direction,
So total force acting on the body will be $F_t=F_3-$(resultant of $F_{1}and{F}_2$)

i.e. $F_t=\sqrt{2}F-F=F(\sqrt{2}-1)$

So the acceleration on the body due to these three forces will be

$a_f=F_t/m=(\sqrt{2}-1)F/m=(\sqrt{2}-1)ma/m=(\sqrt{2}-1)a$ .