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A uniform ladder of mass 10kg leans against a smooth vertical wall making an angle 53^(0) 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. |
Answer» Solution : The LADDER is in equilibrium `:.N_(1)=f` and `N_(2)=W` TAKING torque about .B. `N_(1)(AO)=W(CB)`. or `N_(1)(AB)cos53=W((AB)/(2))sin53` or `N_(1)=(2)/(3)W` and `N_(2)=W=10xx9.8=98N` The FRICTIONAL force is `f=N_(1)=(2)/(3)W=65N` |
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