Question

A uniform ladder of mass $15\text{kg}$ leans against a smooth vertical wall making an angle of ${37}^{\circ }$ with it. The other end rests on a rough horizontal floor. Which of the following is/are correct?

• A. Normal force by the floor on the ladder is $150\text{N}$.
• B. Normal force by the floor on the ladder is $170\text{N}$.
• C. Frictional force by the floor on the ladder is $56.25\text{N}$.
• D. Frictional force by the floor on the ladder is $60\text{N}$.

Answer 1

Answer: (A), (C)

Frictional force by the floor on the ladder is $56.25\text{N}$.

Let us suppose normal force and frictional force by the floor on the ladder are $N$ and $f$ respectively as shown in the figure.


For translational equilibrium along vertical,
$\sum F_y=0$
$\Rightarrow \text{Normal force}\left(N\right)\text{= Weight of the ladder}\left(mg\right)$
$\therefore N=15\times 10=150\text{N}$
For translational equilibrium along horizontal,
$\sum F_x=0$
$\Rightarrow f=N^{\prime }$ ... (1)
For rotational equilibrium about point $A$,
$\sum \tau =0$
$\Rightarrow 0\times N-0\times f+150\times 2.5\cos {53}^{\circ }-N^{\prime }\times 5\sin {53}^{\circ }=0$
[taking clockwise positive and $\tau =r_{\perp }F]$
$\Rightarrow 225-4N^{\prime }=0$
$\Rightarrow N^{\prime }=56.25\text{N}$
$\Rightarrow f=56.25\text{N}$ [from (1)]
Hence, options (a) and (c) are the correct answers.