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$A\, 5\, m$ long pole of $3\, kg$ mass is placed against a smooth vertical well as shown in the figure. Under equilibrium condition, if the pole makes an angle of $37^o$ with the horizontal, the frictional force between the pole and horizontal surface is
$20\, N$
$30\, N$
$20\, \mu N$
$30\, \mu N$
Solution
Here, $\theta=37^{\circ}$.
Pole is in translational and rotational equilibrium.
So, for translational equilibrium, after balancing the vertical and horizontal forces, $m g=N_2$ and $f=N_1$.
Here, $N_1$ and $N_2$ are normal reaction forces and $f$ is the frictional force.
The balancing torques about $A$ is, $m g \frac{O A}{2}=N_1 O B \quad \ldots$ (i)
(Torque of $N_2$ and $f$ is zero as they are passing from point $A$, so perpendicular distances are zero)
$O A=5 \cos 37^{\circ}=5 \frac{4}{5}=4 m$.
$O B=5 \quad \sin 37^{\circ}=5 \quad \frac{3}{5}=3 m$.
So, from equation (i),
$m g \frac{4}{2}=N_1(3)$
$\Rightarrow N_1=\frac{2}{3}(3)(10)=20\; N$
$m=3\; kg$
Therefore, the frictional force is, $f=N_1=20\; N$.
