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A uniform ladder of mass 10 kg leans against a smooth vertical wall making an angle 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. |
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Answer» Solution :The LADDER is in EQUILIBRIUM. `THEREFORE N_(1)=f` and `N_(2)=W` Taking torque about .B. `N_(1)(AO)=W(CB)`. or `N_(1)(AB) cos 53 = W((AB)/(2))sin 53` or `N_(1)=(2)/(3)W` and `N_(2)=W=10xx9.8=98 N`. The frictional force is `f=N_(1)=(2)/(3)W=65 N`
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