A uniform wooden stick of mass 1.6 kg and length l

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Problem: A uniform wooden stick of mass 1.6 kg and length $l$ rests in an inclined manner on a smooth, vertical wall of height $h\,( < l)$ such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30 degree with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio $h/l$ and the frictional force $f$ at the bottom of the stick are (g = 10 m/s2)  (IIT JEE 2016)

  1. $\frac{h}{l}=\frac{\sqrt{3}}{16}$, $f=\frac{16\sqrt{3}}{3}$ N
  2. $\frac{h}{l}=\frac{{3}}{16}$, $f=\frac{16\sqrt{3}}{3}$ N
  3. $\frac{h}{l}=\frac{3\sqrt{3}}{16}$, $f=\frac{8\sqrt{3}}{3}$ N
  4. $\frac{h}{l}=\frac{3\sqrt{3}}{16}$, $f=\frac{16\sqrt{3}}{3}$ N

Solution: The forces acting on the stick are its weight $mg$ at the centre of mass $C$, normal reaction $N_A$ at the contact point A due to the floor, frictional force $f$ at the point A due to the rough floor, and the normal reaction $N_B$ at the contact point B due to the wall. Since stick is uniform, its centre of mass C lies at the middle point i.e., at a distance $l/2$ from the end A. It is given that the direction of $N_B$ is perpendicular to the stick and $N_B=N_A$. Resolve $N_B$ in the horizontal and the vertical directions.

A uniform wooden stick of mass 1.6 kg

Since the stick is in equilibrium, the net forces on the stick in the horizontal and the vertical directions are zero i.e., \begin{align} \label{eqn:jzb:1} &N_A+N_B\sin30-mg=0, \\ \label{eqn:jzb:2} &N_B\cos30-f=0. \end{align} Also, in equilibrium, net torque on the stick about any point should be zero. The net torque on the stick about the point A is \begin{align} \label{eqn:jzb:3} mg (l/2)\cos60-N_B (h/\cos30)=0. \end{align} Substitute $N_B=N_A$ and solve above equations to get \begin{align} \frac{h}{l}=\frac{3\sqrt{3}}{16} \end{align} and \begin{align} f=\frac{16\sqrt{3}}{3}\,\mathrm{N}.\nonumber \end{align} We encourage you to solve this problem by taking the torque about the point B or C.

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