The frictional force acting on the block if it does not slide relative to the wedge which moves with an acceleration a towards right is

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Friction $f$ and the component of pseudo force ($ma$) act up the incline and down the incline $mg\sin\theta$ Is acting, which gives $f+ma\cos\theta=mg\sin\theta$ which gives $f$ as $m(g\sin\theta-a\cos\theta)$, but the answer is given as $m(a\cos\theta-g\sin\theta)$.

Please tell me what am I doing wrong.


closed as off-topic by garyp, ZeroTheHero, Bill N, Kyle Kanos, Jon Custer Sep 4 '17 at 16:13

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$f$ does not act up the incline but down the incline. The sign is opposite and then the whole thing fits.

Remember that the wedge is trying to move away from the block towards the right. The static friction force is trying to pull the box along with the wedge towards the right to make it follow.


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