Question

A light rod AB of length 30 cm. rests on two pegs 15 cm. apart. At what distance from the end A the pegs should be placed so that the reaction of pegs may be equal when weight 5W and 3W are suspended from A and B respectively

• A. 1.75 cm., 15.75 cm.
• B. 2.75 cm., 17.75 cm.
• C. 3.75 cm., 18.75 cm.
• D. None of these

Answer 1

Answer: (C)

3.75 cm., 18.75 cm.

Answer 2

Let R, R be the reactions at the pegs P and Q such that AP = x

Resolving all forces vertically, we get

$R + R = 8 W \Rightarrow R = 4 W$

Take moment of forces about A, we get

$R \cdot A P + R \cdot A Q = 3 W \cdot A B$

$\Rightarrow 4 W \cdot x + 4 W \cdot ( x + 15 ) = 3 W \cdot 30$

$\Rightarrow x = 3.75 \mathrm {~cm}$

$\therefore A P = x = 3.75 \mathrm {~cm}$ and $A Q = 18.75 \mathrm {~cm}$