Question

A massless uniform rod is subjected to force $F$ at its free end as shown in figure.

The ratio of tensile stress at plane $P_1$ to stress at $P_2$ is

• A. $1:2$
• B. $\sqrt{2}:1$
• C. $1:4$
• D. $3:2$

Answer 1

The correct option is A $1:2$

As the rod is at rest. So net external force on rod must be zero, therefore a reaction force $F$ from roof is acting vertically upward to cancel the effect of applied $F$ acting vertically downward.
Therefore at any cross section we cut we will have a force $F$ acting vertically on that section which will be cancelled by applied $F$ for lower cross section or by $F$ reaction force from roof for upper cross section.
Lets take upper section $P_1$ where stress is acting perpendicular to it. A Force $F$will act verically downward. Since the cross section on which stress is acting is at 60° to the vertical we need to take component of $F$ in that direction.
⇒ Tensile stress at $P_1=\frac{Fcos60°}{A}=\frac{F}{2A}$
Similarly, Tensile stress at $P_2=\frac{F}{A}$
⇒ $\frac{Tensilestressat{P}_1}{Tensilestressat{P}_2}=\frac{\frac{F}{2A}}{\frac{F}{A}}=\frac{1}{2}=1:2$.