# Static & Kinetic Friction + An Applied Force on an Inclined Ramp

• slybuster
In summary, for a box with a mass of 22 kg at rest on a ramp inclined at 45 degrees to the horizontal, with coefficients of friction μs = 0.78 and μk = 0.65, the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest is 4.3*10N. The magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest is 42.999N. In both parts of the problem, the force needed to overcome static friction is essential in keeping the box at rest.
slybuster

## Homework Statement

A box with a mass of 22 kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficients of friction between the box and the ramp are μs = 0.78 and μk = 0.65

a) Determine magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.

b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.

## Homework Equations

ΣF = Fμs (or Fμk) +/- Fax (or Fay)
ΣF = 0

## The Attempt at a Solution

When I diagram the problem, it seems to me that the block should be in motion (as the horizontal component of gravity [Fgx = Fg*sin(45)]is greater than the static friction inherent to the system [Ffμs = Fn*μs]), but the problem states the the system is at rest.
The following are my attempts to answer the question, can you please tell me if I am misunderstanding static & kinetic friction and getting the wrong answer?

a)
m = 22 kg
θ = 45
μs = .78
g = 9.8 m/s^2
***Fax = ?

Fa = Fgx + Fμs
Fa = (mg*sinθ) + (mg*cosθ*μs)
Fa = 152.45N + 118.91N
Fa = 2.7*10^2N

b)
For this part of the problem, I tried to include μk (i.e. confusing because the problem states the block starts from rest?!)

m = 22 kg
θ = 45
μs = .78
μk = .65
g = 9.8 m/s^2
***Fay = ?

Fx = μk*mg*sinθ
Fx = 99.09N (Down)

μs(Fgy + Fay) = Fx
Fay(0.78) = 19.5N
Fay = 2.5*10N

I feel like this is totally wrong because both kinetic and static friction are involved in the problem...can someone please look at my answers and see if I'm on the right track and have the correct answers? Thank you in advance for reading this...

Part a - fine.
Part b - No additional force along x direction, just in y pushing down onto the block.
This additional force need to be such that the box will not slide down under its own weight, so there is no motion involved.

Apologies, but I don't understand why this is so...what happens to the force of gravity directed in the x direction? It is large enough that it overcomes the static friction needed to move the block...is the question just needlessly inserting a μk?

I understand that if the system magically appears with the force on top of it, then the block would not move...but the whole part in the question stating that the block starts at rest alongside a μk coefficient being included leads me to believe that they want me to consider the system dynamic--and not just insert an extra force into the initial constraints.

I'm guessing that it's just a poorly worded question though--wouldn't be the first time in this course...it's a little silly to insert a red herring into an independent study course though.

Wait, perhaps the μk is included in order to be added to the original question? But that would be along the same logic I followed for part B...it's just very strange that the question includes kinetic friction but describes the system as being at rest...perhaps if they had said constant velocity it would make more sense...but then you'd just need kinetic and not static though; correct?

Sorry, independent study...trying to milk/learn as much as I can from the sparse materials provided. This forum is just great and I really appreciate the help I'm getting.

slybuster said:
Wait, perhaps the μk is included in order to be added to the original question? But that would be along the same logic I followed for part B...it's just very strange that the question includes kinetic friction but describes the system as being at rest...perhaps if they had said constant velocity it would make more sense...but then you'd just need kinetic and not static though; correct?

Sorry, independent study...trying to milk/learn as much as I can from the sparse materials provided. This forum is just great and I really appreciate the help I'm getting.
This is unfortunately a badly worded problem for reasons you have already noted. I guess ignore the initial 'at rest' wording and your answer to part a is correct. For part b, proceed as suggested by Basic-Physics in post #2. The kinetic friction coefficient uk is not needed.

In part b the block is at rest so one need to consider just static friction. Without any additional force pushing down onto the block it would have slided down under its own weight. So the question is what minimum force will keep it at rest then. Pushing down onto the block would increase the normal force and thus the static friction which need to balance the x component of the weight then.

Alright...so it's just

μs(Fgy + Fa) = Fgx
μs(mg*cosθ+Fa) = mg*sinθ
.78*Fa = 33.539N
Fa = 42.999N

Fa = 4.3*10N

Correct? I noted that because the problem states that the block starts from rest, the applied force should be interpreted as an initial condition inherent to they system being described.

slybuster said:
Alright...so it's just

μs(Fgy + Fa) = Fgx
μs(mg*cosθ+Fa) = mg*sinθ
.78*Fa = 33.539N
Fa = 42.999N

Fa = 4.3*10N

Correct? I noted that because the problem states that the block starts from rest, the applied force should be interpreted as an initial condition inherent to they system being described.

Not following you , but which part have you done. Part a or b ?

Well I give you hints with hidden concepts : (As PhanthomJay notes , this problem is badly worded for sure) In first part , your applied force is largest if it overcomes both friction and gravity.
In second part , your force perpendicular to block must be such that the out coming normal reaction constitutes static friction so that gravitational force is just balanced.

I just noted that you got part a.

slybuster said:
Alright...so it's just

μs(Fgy + Fa) = Fgx
μs(mg*cosθ+Fa) = I had it plugged into the outlet bar by the tv... Middle room. *sinθ
.78*Fa = 33t.539N
Fa = 42.999N

Fa = 4.3*10N

Correct? I noted that because the problem states that the block starts from rest, the applied force should be interpreted as an initial condition inherent to they system being described.
Yes, but as I mentioned, the
wording of the problem is not very good. In order for the box to be initially at rest, there must be an inherent upwardly applied force parallel to the incline of sufficient magnitude to keep it from sliding down the plane ( and not large enough to make it slide up the plane, per your calc in part a), OR, there must be an inherent downward applied force perpendicular to the plane of sufficient magnitude to keep it from sliding down and of any magnitude greater than that, OR, some combination thereof. In part a you assume the initial parallel force, and part b you assume the initial perpendicular force, so it becomes 2 distinct problems. Better to forget the initial at rest statement. Problem should have been worded differently.

I'm becoming more confused...

The first attempt at b (located in post 1) attempted to take an initial motion into account but I am unsure if my solution/breakdown of the problem makes sense. (?)

The second attempt at b (located in post 7) attempts to establish the force as an initial condition.

Are both solutions carried out correctly? If the static friction (μs) is broken, does it take a smaller force to restore it, relative to the magnitude of the force required to prevent the object overcoming the μs, if kinetic friction (μk) is taken into account?

Apologies, I thank you for all of your help. This ILC course is a nightmare...

slybuster said:
μs(Fgy + Fa) = Fgx
μs(mg*cosθ+Fa) = mg*sinθ
the problem states that the block starts from rest, the applied force should be interpreted as an initial condition inherent to they system being described.
That looks right to me.
I feel that the point of the 'initially at rest' info is so that you know to use μs, not μk. It reads strangely because you naturally think in terms of there being a time lag, however short, before the normal force is applied.

Fa = 4.3*10N √
You did good.
Quite a tricky problem and one can easily get it wrong.

Thank you so much everyone. I really don't know what I would do without this site...I think I'll purchase a membership.

What a hobby you guys have! Helping those in need while keeping your mind sharp--I don't know what could be better.

Thanks again.

Remember the maths is just a tool, the physics dictates how you should use this tool.
This means that you need to try and get a feeling for the underlying physics in the problem - what is happening with the physics - before applying the maths to it.

## 1. What is static friction?

Static friction is the force that prevents an object from moving when a force is applied to it. It occurs when the surfaces of two objects are in contact with each other and there is no motion between them.

## 2. How is static friction different from kinetic friction?

Static friction only occurs when an object is at rest, whereas kinetic friction occurs when an object is in motion. Additionally, static friction is typically greater than kinetic friction, meaning it requires more force to overcome static friction and start an object in motion.

## 3. How is an applied force on an inclined ramp related to friction?

When an object is placed on an inclined ramp, the force of gravity acts on the object and pulls it down the ramp. The normal force, which is perpendicular to the ramp, counteracts the force of gravity. Friction is then applied in the opposite direction of the motion, preventing the object from sliding down the ramp too quickly.

## 4. How does the angle of the ramp affect the friction force?

The angle of the ramp affects the normal force and therefore the friction force. As the angle of the ramp increases, the normal force decreases and the friction force decreases as well. This means that the object will be more likely to slide down the ramp at steeper angles.

## 5. What factors can affect the amount of friction on an inclined ramp?

The amount of friction on an inclined ramp can be affected by the roughness of the surfaces in contact, the weight of the object, and the angle of the ramp. Other factors such as temperature and the presence of lubricants can also affect friction. Additionally, the coefficient of friction, which is a measure of the roughness of the surfaces, can greatly impact the amount of friction present.

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