Question

A rod of length L and mass M is acted on by two unequal forces $F_1$ and $F_2(<F_1)$ as shown in the figure. The tension in the rod at a distance y from the end A is given by:

• A. $F_1[1-\frac{y}{L}]+F_2\left[\frac{y}{L}\right]$
• B. $F_2[1-\frac{y}{L}]+F_1\left[\frac{y}{L}\right]$
• C. $(F_1-F_2)\frac{y}{L}$
• D. None of these

Answer 1

The correct option is A

$F_1[1-\frac{y}{L}]+F_2\left[\frac{y}{L}\right]$

Net force on the rod $=F_1-F_2$
Acceleration of rod $=\frac{F_1-F_2}{M}$
Rod of length ‘L’ has mass = M
Rod of length ‘y’ has mass $=\frac{M}{L}y$
$F_1-T=\frac{M}{L}y.a{F}_1-T=\frac{M}{L}y\frac{F_1-F_2}{M}T=F_1[1-\frac{y}{L}]+F_2\frac{y}{L}$