Question

A crate of mass 100.0 kg rests on a rough surface inclined at an angle of 37.0° with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is 80% of that for the static case a. What is the coefficient of static friction? b. What is the maximum force that can be applied upward along the plane on the rope and not move the block? c. With a slightly greater applied force, the block will slide up the plane. Once it begins to move, what is its acceleration and what reduced force is necessary to keep it moving upward at constant speed? d. If the block is given a slight nudge to get it started down the plane, what will be its acceleration in that direction? e. Once the block begins to slide downward, what upward force on the rope is required to keep the block from accelerating downward?

          A crate of mass 100.0 kg rests on a rough surface inclined at an
angle of 37.0° with the horizontal. A massless rope to which a
force can be applied parallel to the surface is attached to the
crate and leads to the top of the incline. In its present state,
the crate is just ready to slip and start to move down the plane.
The coefficient of friction is 80% of that for the static case
a.    What is the coefficient of static
friction?
b.    What is the maximum force that can be
applied upward along the plane on the rope and not move the
block?
c.     With a slightly greater applied
force, the block will slide up the plane. Once it begins to move,
what is its acceleration and what reduced force is necessary to
keep it moving upward at constant speed?
d.    If the block is given a slight nudge
to get it started down the plane, what will be its acceleration in
that direction?
e.    Once the block begins to slide
downward, what upward force on the rope is required to keep the
block from accelerating downward?

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Added by Andrea B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ivan Kochetkov

10/04/2022

Step 1

The crate is just ready to slip and start to move down the plane, which means the force of static friction is equal to the component of the weight of the crate acting down the plane. The weight of the crate is mg, where m is the mass of the crate and g is the Show more…

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A crate of mass 100.0 kg rests on a rough surface inclined at an angle of 37.0° with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is 80% of that for the static case a. What is the coefficient of static friction? b. What is the maximum force that can be applied upward along the plane on the rope and not move the block? c. With a slightly greater applied force, the block will slide up the plane. Once it begins to move, what is its acceleration and what reduced force is necessary to keep it moving upward at constant speed? d. If the block is given a slight nudge to get it started down the plane, what will be its acceleration in that direction? e. Once the block begins to slide downward, what upward force on the rope is required to keep the block from accelerating downward?

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Transcript

00:01 Hello, we have to solve the following problem.

00:02 A crate has a mass of 100 kilograms and it rests on a rough surface inclined at an angle of 37 degree with the horizontal.

00:14 A massless slope to which a force can be applied is connected parallel to the surface and it leads to the top of the incline.

00:23 So let's catch it.

00:26 Here this crate experiences the gravity downwards and normal reaction perpendicular to the incline let's also introduce the axis y will be perpendicular and x -axis will be up the incline here coefficient of kinetic friction is 80 % of the one of static friction and the crate is about to sleep and start to move down the plane so it means that it's better for us to change the direction of this x -axis let's direct it down the plane and it means that here the static force is acting in the first question static force is acting up the incline we have to first determine coefficient of static friction let's do this so here static friction force in the first case equals to m g sine alpha and that is and the static friction force is also coefficient of static friction force times normal reaction which is coefficient of static friction force mj cosine alpha thereby sine alpha is coefficient of static friction cosine alpha and that's why coefficient of static friction is tangent of 30 degree let's calculate it that is zero 500 0 .58 let's move on.

02:42 In question b, we have to find the maximum force which can be applied upwards along the ropes of the block doesn't move.

02:56 So this maximum force equals to the sum of gravity, or basically it's x component and static friction force.

03:59 That is 980 newton's.

04:05 Now let's move on to the next question question c.

04:10 In question c, we have to calculate acceleration and the force and the reduced force which is needed to run it at a constant speed.

04:31 So let's split it into two sub questions.

04:34 First we will calculate acceleration...

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