If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is θ.

Answers

Answer 1

Answer:

The maximum angle of an incline above the horizontal for which an object will not slide down is,

$\theta=\tan^(-1)\mu$

What is friction force?

Friction force is the force which resist the motion of a body, when the the body is in contact with another body or surface.

The friction force of a body is the product of normal force acting on the body and the coefficient of the friction of that body. It can be given as,

$F_f=F_n*\mu$

Here, $F_n\n$ is the normal force and $\mu$ is the coefficient of the friction.

If a force is at the rest, then the net force acting on a body should be equal to zero. Now the normal force acting on a body is product of mass and the gravity.

For the inclined plane the normal force can be given as,

$F_n=mg\cos\theta$

Put this value in the above equation as,

$F_f=mg\cos\theta*\mu$

Now for the horizontal component of velocity to balance the object, is given as,

$F_f=mg\sin\theta$ .

Comparing both the values of friction force, the equation become,

$mg\sin\theta=mg\cos\theta*\mu\n\sin\theta=\cos\theta*\mu\n(\sin\theta)/(\cos\theta)=\mu\n\tan\theta=\mu\n\theta=\tan^(-1)\mu$

Thus, the maximum angle of an incline above the horizontal for which an object will not slide down is,

$\theta=\tan^(-1)\mu$

Learn more friction force here;

brainly.com/question/13680415

Answer 2

Answer:

Answer:

$\theta = tan^(-1)\mu$

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

$F_f = mgsin\theta$

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

$F_f = \mu N$

$N = mg cos\theta$

so we have

$F_f = \mu (mg cos\theta)$

so we will have

$mg sin\theta = \mu (mg cos\theta)$

so now we have

$tan\theta = \mu$

so maximum possible angle of the inclined plane is

$\theta = tan^(-1)\mu$

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D. distance.

Answers

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With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock?
Answer in units of $m / s$.

Answers

Answer:

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Plugging in the numbers, we get:

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Since the cliff divers want to clear the rock, they need to travel a horizontal distance of at least $9.34 m$ during this time. We can use the equation for horizontal motion again to solve for the minimum initial velocity:

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Therefore, the minimum horizontal velocity that the cliff divers must have to clear the rock is approximately $6.49 m/s$.

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Answers

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Answers

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