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Question

At the bottom edge of a smooth wall, an inclined plane is kept at an angle of $${45}^{o}$$. A uniform ladder of length $$l$$ and mass $$M$$ rests on the inclined plane against the wall such that it is perpendicular to the incline. 
(a) If the plane is also smooth, in which direction will the ladder slide?
(b) What is the minimum coefficient of friction necessary so that the ladder does not slip on the incline.


Solution

The ladder has tendency to slip by rotating clockwise about the point $$A$$
(b) The free-body diagram of the ladder is shown in figure.
Balancing torque about point $$ fl=Mg\left( \cfrac { l }{ 2 } \sin { { 45 }^{ o } }  \right) $$
$$\therefore f=\cfrac { Mg }{ 2\sqrt { 2 }  } $$
Balancing forces in the vertical direction,
$$Mg={ R }_{ 2 }\cos { { 45 }^{ o } } +f\sin { { 45 }^{ o } } ....(ii)\quad $$
From Eqs(i) and (ii) we get
$${ R }_{ 2 }=\cfrac { 3Mg }{ 2\sqrt { 2 }  } .....(iii)$$
From Eqs(i) and (iii), $$f=\mu {R}_{2}$$
or $$\cfrac { Mg }{ 2\sqrt { 2 }  } =\cfrac { 3Mg }{ 2\sqrt { 2 }  } \mu \quad or\quad \mu =\cfrac { 1 }{ 3 } $$
1047440_981986_ans_4ee34fbb8d3c4f4ea1ddc0c0c658b7b2.png

Physics

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