Question 247: Laws of Motion

Question

A block of mass m is placed on a smooth wedge of inclination $$	heta $$. The whole system is accelerated horizontally so that the block  does not slip on the wedge. The force exerted by the wedge on the block will be (g is acceleration due to gravity)

Accepted Answer

The wedge is given an acceleration to the
left.
$$ 	herefore $$ The block has a
pseudo acceleration to
the right, pressing
against the wedge
because of which the
block is not moving.
$$ 	herefore $$  mgsin$$	heta $$ = macos$$	heta $$
or $$a = {{g\sin 	heta } \over {\cos 	heta }}$$
Total reaction of the wedge on the block is
N = mgcos$$	heta $$ + masin$$	heta $$.
$$ \Rightarrow N = mg\cos 	heta  + {{mg\sin 	heta \sin 	heta } \over {\cos 	heta }}$$
$$ \Rightarrow N = {{mg\left( {{{\cos }^2}	heta  + {{\sin }^2}	heta } ight)} \over {\cos 	heta }} = {{mg} \over {\cos 	heta }}$$

Answer

mg cos$$	heta $$

Answer

mg sin$$	heta $$

Answer

mg

Answer

mg/cos $$	heta $$

Laws of Motion

A block of mass m is placed on a smooth wedge of inclination $\theta $. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block will be (g is acceleration due to gravity)

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