Block on Incline

A block of mass M M rests on an fixed incline plane of angle θ \theta . A horizontal force of magnitude M g Mg is applied on the block. If the friction force is large enough to keep the block at rest, for what value of θ \theta in degrees is the normal force exerted by the plane on the block at its maximum?


The answer is 45.

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2 solutions

Discussions for this problem are now closed

Satvik Pandey
Jun 18, 2014

When we resolve weight and applied force perpendicular to the plane we get normal reaction:

N = m g ( sin θ + cos θ ) N = mg(\sin\theta + \cos\theta)

For finding the value of θ \theta for which N N is maximum, we use calculus. Differentiating N N with respect to θ \theta , we get:

d N d θ = m g ( cos θ sin θ ) \frac{dN}{d\theta} = mg(\cos\theta - \sin\theta)

Equating this to 0 0 , we get: θ m a x = 45 \theta_{max} = 45 degrees

Nice explanation... It's the best answer...

Kumar Manoj Kalita - 6 years, 11 months ago

You can also convert the expression to a single sine term and maximise it,

An observation: The question doesn't mention whether the inclined plane is rough or smooth.

If it is smooth θ = π 4 \theta=\frac{\pi}{4} is the only solution (by balancing forces along the incline) and the normal force has a unique value of N = 2 m g N=\sqrt{2}mg

Soumen Goswami - 6 years, 7 months ago

The maximum reaction of plane to horizontal force can only be achieved with 45 degree incline plane. A physics fact.

Proof please?

Jayakumar Krishnan - 6 years, 12 months ago

Normal reaction here is mg(cosA + sinA) which is maximum when A=45

faizal shaikh - 6 years, 12 months ago

using trigononmetric identities was a better option coz i dont know calculas...

Faraz Mohammed - 6 years, 9 months ago

Thanks satvik

Sarang Zende - 6 years, 11 months ago

You are welcome!

satvik pandey - 6 years, 8 months ago

Value of friction cofficient???????

Gajendra Singh Chauhan - 6 years, 10 months ago

There is no friction.

satvik pandey - 6 years, 8 months ago

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