Question

A rod of mass m is supported on a wedge of mass M as shown in figure. Find the accelerations of rod a and wedge A in the arrangement. Assume that the friction between all contact surfaces is negligible.

• A. $A=\frac{Mg}{(mtan\alpha +Mcot\alpha )}$
• B. $A=\frac{Mg}{(msin\alpha +Mcot\alpha )}$
• C. $a=\frac{mgtan\alpha }{(mtan\alpha +mcot\alpha )}$
• D. $a=\frac{Mgtan\alpha }{(Msin\alpha +Mcot\alpha )}$

Answer 1

The correct option is A

$A=\frac{Mg}{(mtan\alpha +Mcot\alpha )}$

Free body drawing of

Since horizontal forces are irrelevant,

mg - N cos $\alpha$ = ma ...(i)

Free body drawing of wedge

Vertical forces are irrelevant

∴ N sin $\alpha$ = MA ....(2)

Constraint relation is a = A tan $\alpha$ ...(3)

We have to find a and A.

$\therefore$ Substituting (2) in (1)

$mg-\frac{MA}{sin\alpha }xcos\alpha =ma$

$\Rightarrow mg-MAcot\alpha =ma$

$\Rightarrow ma+MAcot\alpha =mg$ -----------(4)

Solving (4) and (3)

$mAtan\alpha +MAcot\alpha =mg$

$\Rightarrow A=\frac{mg}{(mtan\alpha +Mcot\alpha )}$

$a=\frac{mgtan\alpha }{(mtan\alpha +Mcot\alpha )}$