Question

A uniform ladder of mass $10kg$ leans against a smooth vertical wall making an angle of $53$ with it. The other end rests on a rough horizontal floor. Find the normal force and the frictional force that the floor exerts on the ladder

• A. $65N,65N,98N$
• B. $65N,98N,65N$
• C. $98N,65N,65N$
• D. $65N,65N,65N$

Answer 1

The correct option is B $65N,98N,65N$

$Given$ :- Mass of ladder $m$ = 10 kg

Let, Acceleration due to gravity $\left(g\right)$ = 9.8 ${m/s}^2$

Angle between ladder and wall = 53°

$ToFind$ :- $1$. Normal forces $\left(N_{1}and{N}_2\right)$

$2$. Frictional force $\left(f\right)$

$Solution$ :- since, the ladder is in balanced condition

$\therefore N_1=f$ $(N_1\leftrightarrows f)$

And, $N_2=W=mg$ $(N_2\uparrow \downarrow W)$

Now, finding Torque on ladder about A (53°) we get,

$N_1\left(AB\right)cos53°=W\left(\frac{AB}{2}\right)sin53°$

$N_1\times \frac{3}{5}=\frac{mg}{2}\times \frac{4}{5}$

$\therefore$ $N_1=\frac{98\times 2}{3}⟹\underset{–}{65N}$

Now, $N_2=mg⟹\underset{–}{98N}$

And, $N_1=f⟹\underset{–}{65N}$

Hence , $\underset{–}{OptionB(65N,98N,65N)iscorrect.}$