Question

On a rod of length $L$ & mass $M$, how much tension will be at a distance $y$ from $F_1$ when two dissimilar forces $F_1$ & $F_2(F_2<F_1)$ are applied on the rod?

• A. $F_1(1-\frac{y}{L})+F_2\left(\frac{y}{L}\right)$
• B. $\frac{M}{L}y+\left(\frac{F_1-F_2}{M}\right)$
• C. $F_1(1+\frac{y}{L})+F_2\left(\frac{y}{L}\right)$
• D. $\frac{M}{L}y+\left(\frac{F_1+F_2}{M}\right)$

Answer 1

Answer: (A)

$F_1(1-\frac{y}{L})+F_2\left(\frac{y}{L}\right)$

Considering the part $BC$

$F_2-T=\frac{m(L-y)a}{L}$ ( $L⟶$M,$y⟶$$\frac{My}{L}$)

considering the part $AB$

$T-F_1=\frac{Mya}{L}$ ($L⟶$M,$(L-y)⟶$$\frac{M(L-y)}{L}$)

dividing the above equations and rearranging we get,

$T=\frac{y{F}_2}{L}+\frac{(L-y)F_1}{L}$

So,option $A$ is correct.