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
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.
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