Question

A rod of length $L$ and mass $M$ is acted on by two unequal forces $F_1$ and $F_2\left(<F_1\right)$ as shown in the figure.


The tension in the rodata distance $y$ from the end $A$ is given by

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

Answer 1

Answer: (A)

Net force on the $\mathrm{rod}=F_1-F_2$
As mass of the rod is $M,$ so acceleration of the rod is
$a=\frac{\left(F_1-F_2\right)}{M}$
Now considering the motion of part $AB$ of the rod [whose mass
is equal to $(M/L)y]$, then the equation of motion is
$F_1-T=\frac{M}{L}y\times a$
where $T$ is the tension in the rod at $B$.
or $F_1-T=\frac{M}{L}y\times \left(\frac{F_1-F_2}{M}\right)(Using(i))$
or $T=F_1\left(1-\frac{y}{L}\right)+F_2\left(\frac{y}{L}\right)$
Second method: To calculate tension at $B$ we can also consider the motion of the other part of the rod $i.e.,BC.$ Then the equation of motion will be
$T-F_2=\frac{M}{L}(L-y)\times a$
or $T=F_2+\frac{M}{L}(L-y)\times \frac{\left(F_1-F_2\right)}{M}(Using(i))$
or $T=F_1\left(1-\frac{y}{L}\right)+F_2\left(\frac{y}{L}\right)$