• # question_answer                 A weightless ladder 20 ft long rests against a frictionless wall at an angle of  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: 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 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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