Question

Three blocks of masses $m , 3\, m$ and $5\, m$ 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 $16\, N$ 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)

8 N, 2 N

Answer 2

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