Question

Three blocks of masses $m,3m$ and $5m$ are connected by massless strings and pulled by a force $F$ on a frictionless surface as shown in the figure below. The tension $P$ in the first string is $16N$ If the point of application of $F$ is changed as given below
The value of $P$ and $Q$ shall be

• A. 16 N, 10 N
• B. 10 N, 16 N
• C. 8 N, 2 N
• D. 10 N, 6 N

Answer 1

Answer: (C)

Let a be the common acceleration of the system
$\therefore a=\frac{F}{m+3m+5m}=\frac{F}{9m}\dots .$ (i)
$\therefore P=(3m+5m)a$
$16=8ma(\because P=16N(\text{ Given })]$
$a=\frac{2}{m}$
Substituting this value of a in eqn(i), we get $F=18N$