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 |
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