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. F₁$\left( 1 - \frac{y}{L} \right) + F_{2}\left( \frac{y}{L} \right)$
• B. $F_{2}\left( 1 - \frac{y}{L} \right) + F_{1}\left( \frac{y}{L} \right)$
• C. (F₁ – F₂) $\frac{y}{L}$
• D. None of these

Answer 1

Answer: (A)

F₁$\left( 1 - \frac{y}{L} \right) + F_{2}\left( \frac{y}{L} \right)$

Answer 2

a = $\frac{F_{1} - F_{2}}{m_{1} + m_{2}} = \frac{F_{1} - F_{2}}{M}$

and

F₁ – T = m­₁a

or F₁ – T = $\frac{M}{L}.y.\frac{(F_{1} - F_{2})}{M}$

or F₁ – T = (F₁ – F₂) $\frac{y}{L}$

or T = F₁ – $\frac{(F_{1} - F_{2})y}{L}$

or T = $\frac{F_{1}(L - y)}{L} + \frac{F_{2}y}{L}$