question_answer

A weightless ladder 20 ft long rests against a frictionless wall at an angle of \[{{60}^{o}}\] from the horizontal. A 150 pound man is 4 ft from the top of the ladder. A horizontal force is needed to keep it from slipping. Choose the correct magnitude of force from the following:                    [AIPMT 1998]

A) 17.3 pound

B) 100 pound        

C) 120 pound 

D) 150 pound

Correct Answer: A

Solution :

Key Idea: The net moment about point of contact between ground and ladder should be zero.

Let (as shown in figure) AB be a ladder and F be the horizontal force to keep it from slipping.

w is the weight of man. Suppose \[{{N}_{1}}\] and \[{{N}_{2}}\] be normal reactions of ground and wall respectively.

In horizontal equilibrium,

\[{{N}_{2}}=F\]

In vertical equilibrium,

\[{{N}_{1}}=w\]

Taking moments about A;

Clockwise torque = Anticlockwise torque

\[{{N}_{1}}\times CD={{N}_{2}}\times OB\]

but in    \[\Delta AOB,\,\sin \,\,{{60}^{o}}\,=\frac{OB}{AB}\]

\[\Rightarrow \]   \[OB=AB\sin {{60}^{o}}\]

In \[\Delta BCD,\]

\[\cos {{60}^{o}}=\frac{CD}{BC}\]

\[\Rightarrow \]   \[CD=BC\,\cos \,{{60}^{o}}\]

Substituting in Eq. (i) , we have

\[{{N}_{1}}\times BC\,\cos \,\,{{60}^{o}}={{N}_{2}}\times AB\,\sin \,{{60}^{o}}\]

\[\Rightarrow \]   \[w\times BC\times \frac{1}{2}=F\times AB\times \frac{\sqrt{3}}{2}\]

Given: \[w=150\] pound, \[AB=20\,ft.\],\[BC=4\text{ }ft\].

\[\therefore \]      \[150\times 4\times \frac{1}{2}=F\times 20\times \frac{\sqrt{3}}{2}\]

\[\Rightarrow \]   \[F=\frac{150\times 4}{20\sqrt{3}}\]

\[=\frac{150\times 4\times \sqrt{3}}{20\times 3}\]

= 17.3 pound