A mass $m$ $i s p l a c e d o n a w e d g e (t r i a n g u l a r b l o c k) o f m a s s$ M $. T h e w e d g e m o v e s o n a s m o o t h h o r i z o n t a l s u r f a c e w h a t s h o u l d b e t h e f o r c e$ F $a p p l i e d o n t h e w e d g e t o t h e r i g h t, s o t h a t$ m$ remains stationary with respect to wedge. (Ignore any friction)

g tan θ

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m g cos θ

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(M + m) g tan θ

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(M + m) g c o s e c θ

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Solution

The correct option is C $(M + m) g tan θ$
From figure, $m g sin θ = m a cos θ$
$\Rightarrow a = g tan θ$
$∴ F o r c e, F = (M + m) a$
$= (M + m) g tan θ$ .

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A mass misplacedonawedge(triangularblock)ofmassM.ThewedgemovesonasmoothhorizontalsurfacewhatshouldbetheforceFappliedonthewedgetotheright,sothatm$ remains stationary with respect to wedge. (Ignore any friction)

The correct option is C (M+m)gtanθFrom figure, mgsinθ=m a cosθ⇒a=gtanθ∴Force,F=(M+m)a=(M+m)gtanθ.

The correct option is C $(M + m) g tan θ$
From figure, $m g sin θ = m a cos θ$
$\Rightarrow a = g tan θ$
$∴ F o r c e, F = (M + m) a$
$= (M + m) g tan θ$ .