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

Updated On: Aug 1, 2022
  • 16 N, 10 N
  • 10 N, 16 N
  • 8 N, 2 N
  • 10 N, 6 N
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The Correct Option is C

Solution and Explanation

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 $F=18 N$
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Concepts Used:

Tension

A force working along the length of a medium, especially if this force is carried by a flexible medium like cable or rope is called tension.  The flexible cords which bear muscle forces to other parts of the body are called tendons.

Net force = 𝐹𝑛𝑒𝑡 = 𝑇−𝑊=0,

where,

T and W are the magnitudes of the tension and weight and their signs indicate a direction, be up-front positive here.

Read More: Tension Formula