Question

A rod of mass m is supported by string AB and friction due to wall. Then friction force on rod due to the wall is: (g = acceleration due to gravity).

• A. mg upwards
• B. mg downwards
• C. $\frac{mg}{2}$upwards
• D. Data insufficient

Answer 1

The correct option is C $\frac{mg}{2}$upwards

Let friction force be 'f' from the wall and it will act vertically upwards to prevent rod from slipping downwards.

From FBD of rod, balancing the torque about point 'A'.
${\tau }_N=0{(}^{\prime }{N}^{\prime }\text{passes through A})$

${\tau }_T=0\left(\text{Tension passes through A}\right)$

$\Rightarrow {\tau }_f+{\tau }_{mg}=0$

Taking anticlockwise sense of rotation as +ve,

$\Rightarrow +fL-\frac{L}{2}\left(mg\right)=0$

$orfL-\frac{L}{2}mg=0$

$\Rightarrow fL=\frac{L}{2}mg$

$\therefore f=\frac{mg}{2}$

Why this question? Tip: For the rotational equilibrium ${\tau }_{net}=0$ on rod. It will be wise move to choose a reference point for balancing torque about which torque of unknown force vanishes.