• 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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