# Question Concerning The Coefficient of Friction: Block on a Rough Inclined Plane

• UnknownGuy
In summary: Coefficient of Friction= μR In summary, the coefficient of friction between the block and the plane is 0.577.
UnknownGuy

## Homework Statement

A block of mass 20kg is on rest on a rough plane inclined at 60 degrees to the horizontal. A force P is applied to the block to "maintain equilibrium". P is parallel to the line of greatest slope of the inclined plane. The maximum value of P is twice the minimum value of P. Find the coefficient of friction between the block and the plane.

## Homework Equations

Frictional force=Coefficient of friction*Normal Foce (F=μR)

## The Attempt at a Solution

This question can be interpreted in two different ways:
First way: Without P, the block will slide down so P MUST act up the line of greatest slope of the plane.
Second way: P can either act up or down the line of greatest slope of the plane, and the resultant of P, component of the weight down the slope and the frictional force is 0N.
Attached are pictures illustrating the two methods. Can you please tell me which method do you think is correct? In my exam, I was leaning towards the second way but μ=5.20 seemed rather large. Moreover, the wording, "To maintain equilibrium" made it seem like the block would be sliding down without P hence it has to act up the line of greatest slope.

Edit: It seems like the quality of the pictures has been decreased. The subscripts under P are (min) and (max)

#### Attachments

• FirstWay.jpg
20.8 KB · Views: 482
• SecondWay.jpg
20.8 KB · Views: 436
UnknownGuy said:
the wording, "To maintain equilibrium" made it seem like the block would be sliding down without P hence it has to act up the line of greatest slope.
I feel sure that is the intent.

UnknownGuy said:
It seems like the quality of the pictures has been decreased

I don't understand the signs in your firstway picture. For Pmin the friction has to work opposite from the way for Pmax
Check what P has to be in that case: 0=0 is also an equality ...

I don't understand the signs in your second way picture. It looks like you let friction work in the same direction again ...
mu = 5 is nonsense

BvU said:
I don't understand the signs in your firstway picture.
BvU said:
I don't understand the signs in your second way picture.
All looks ok to me.

BvU said:
mu = 5 is nonsense
umm... why?

I will write down the calculations in the pictures here just for clarification.

First picture:
For P(min), the box will be on the point of sliding down with P pushing up the slope.
(1):20gcos(60)μ+P(min)=20gsin(60)
For P(max), the box will be on the point of going up with P pushing up the slope.
(2):P(max)=20gcos(60)μ+20gsin(60)

Setting P(max)=2P(min) and solving (1) and (2) gives μ=0.577

Second picture:
For P(min), the box will be on the point of sliding downwards with P pushing down the slope
(1):P+20gsin(60)=20gcos(60)μ
For P(max), the box will be on the point of sliding up with P pushing up the slope
(2):P=20gcos(60)μ+20gsin(60)

Setting P(max)=2P(min) and solving (1) and (2) gives μ=5.20

Note: I looked it up and found that the largest coefficients of friction never surpass 1.5(ex: dry aluminum on self has μ=1.35) let alone alone 5!
Is this enough proof to regard the second method as incorrect?

Good thing you translated your fuzzy writing: signs are Ok.
You worked out a second scenario for a bonus (my guess is first scenario was intended by exercise composer) -- ##\mu = 5.2 ## is really an awful lot.
It would mean the block would stay in place on a 79 degree slope ( ## \arctan(5.2)\ ## ) -- without glue. But perhaps the term 'nonsense' is too harsh (maybe someone even has an example?).

I withdraw my nonsense. Haru is right ... again !

Ceterum censeo...
Of course ##\ 2\times 0= 0 \ ## so in my book ## \mu = \tan 60^\circ \ = {1\over 2}\sqrt 3\ ## is also a solution but I concede the 'twice' in the problem statement makes it an unlikely candidate.

## 1. What is the coefficient of friction?

The coefficient of friction is a dimensionless quantity that represents the amount of resistance or force between two surfaces in contact. It is typically denoted by the Greek letter mu (μ) and can range from 0 (no friction) to 1 (high friction).

## 2. How is the coefficient of friction calculated?

The coefficient of friction can be calculated by dividing the force of friction (F) by the normal force (N) between the two surfaces: μ = F/N. It can also be determined experimentally by measuring the force required to move an object across a surface at a constant velocity.

## 3. What factors affect the coefficient of friction?

The coefficient of friction can be affected by several factors, including the roughness or texture of the surfaces in contact, the force pressing the surfaces together, the speed of movement, and the presence of any lubricants or contaminants.

## 4. How does the coefficient of friction impact the motion of a block on a rough inclined plane?

The coefficient of friction plays a crucial role in determining the motion of a block on a rough inclined plane. It affects the amount of force required to move the block up or down the incline, as well as the speed and acceleration of the block as it moves.

## 5. How does the angle of the incline impact the coefficient of friction?

The angle of the incline can impact the coefficient of friction in two ways. First, a steeper incline will typically result in a larger normal force, which can increase the coefficient of friction. Second, as the angle of the incline increases, the component of the weight of the block acting parallel to the surface also increases, which can increase the force of friction.

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