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

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1. May 18, 2018

### UnknownGuy

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Frictional force=Coefficient of friction*Normal Foce (F=μR)

3. 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)

#### Attached Files:

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• ###### SecondWay.jpg
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2. May 18, 2018

### haruspex

I feel sure that is the intent.

3. May 18, 2018

### BvU

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

4. May 18, 2018

### haruspex

All looks ok to me.

umm... why?

5. May 18, 2018

### UnknownGuy

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?

6. May 18, 2018

### BvU

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.