Question

A rod of length L and mass M is acted on by two unequal forces F₁ and F₂ (<F₁) as shown in the following figure. The tension in the rod at a 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. $(F_{1} - F_{2})\frac{y}{L}$
• D. None of these

Answer 1

Answer: (A)

$F_{1}\left( 1 - \frac{y}{L} \right) + F_{2}\left( \frac{y}{L} \right)$

Answer 2

 a = $\frac{F_{1} - F_{2}}{M}$

 T = F₁ – m₁a = F₁ – $\left\lbrack \frac{M}{L}.y \right\rbrack\left\lbrack \frac{F_{1} - F_{2}}{M} \right\rbrack$

= $F_{1}\left( 1 - \frac{y}{L} \right) + F_{2}\left( \frac{y}{L} \right)$