FIND THE NORMAL FORCE.......

Block of mass m is stationary with respect to a smooth wedge. Normal force exerted by wedge on the block is

m g t a n θ \frac{mg}{tan\theta} m g c o s θ \frac{mg}{cos\theta} m g cos θ mg\cos\theta m g tan θ mg\tan\theta m g s i n θ \frac{mg}{sin\theta} m g sin θ mg\sin\theta

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1 solution

Forces acting along the inclined plane:

m a ma cos θ \cos\theta = m g mg s i n θ sin\theta

\implies a a = g t a n θ gtan\theta

Forces acting normal to the block:

N N = m g mg c o s θ cos\theta + m a ma s i n θ sin\theta

\implies N N = m g mg c o s θ cos\theta + m( g t a n θ gtan\theta ) s i n θ sin\theta

\implies N N = m g mg c o s θ cos\theta + m g mg s i n 2 θ c o s θ \frac{sin^2\theta}{cos\theta} = m g c o s θ \frac{mg}{cos\theta}

@Harsh Shrivastava good solution.

Anand Badgujar - 3 years, 2 months ago

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Thanks @Anand Badgujar

Harsh Shrivastava - 3 years, 2 months ago

I admire the time you put to make this solution.

Harsh Poonia - 2 years, 3 months ago

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