# Coefficient of Friction of a piano

• Medgirl314
In summary, the men trying to load a 900 kg piano onto a truck lost control and the piano rolled down the loading ramp, losing a significant amount of speed. The ramp is a 12 degree incline that is 3.2m long. If the piano accelerates down the ramp at 0.4m/s^2, find the coefficient of rolling friction.
Medgirl314

## Homework Statement

Two men loading a 900 kg piano onto a truck lose control and the piano slips, rolling down the loading ramp. The ramp is a 12 degree incline that is 3.2m long. If the piano accelerates down the ramp at 0.4m/s^2, find the coefficient of rolling friction.

F=ma

## The Attempt at a Solution

a=g(cos)(theta)-(mu)g(sin)(theta
mu=g(cos)(theta)=9.8(cos12)+0.4
mu=9.9

This looks wrong. Where did I mess up?

Thanks!

Looks like your calculations for weight force and normal force are off, on account of switching the sin and cos. Try drawing it out to see which trig function to use -- weight force acceleration being parallel to the ramp, normal force being normal to it.

Not only is the first equation wrong, as jackarms points out, the second equation does not follow from the first. What happened to the sin, and the sign?

Edit: Too much information

Sketch out the scenario. Think about how you would find the acceleration of the piano if you had its resultant force and the co-efficient of friction given. Then flip the script and find the co-efficient instead. What do you do?

The resultant force Fr is the sum of friction and the slope-directional pull of gravity - what are youmacallit.

Last edited:
Not only is the first equation wrong, as jackarms points out, the second equation does not follow from the first. What happened to the sin, and the sign?

(I think I wrote out a lot more, but didn't want to be annoying, so I didn't include all the steps, just the vital ones. I'll include all of them this time. )

Edit: Too much information.(Ah, I thought that may be the case. My teacher is starting to throw in extra information to trip me up.)

Sketch out the scenario. Think about how you would find the acceleration of the piano if you had its resultant force and the co-efficient of friction given. Then flip the script and find the co-efficient instead. What do you do?
(Okay, I was going off a diagram in my lesson for a nearly identical situation, but it may help to draw a new one)

The resultant force Fr is the sum of friction and the slope-directional pull of gravity - what are youmacallit.

I'll work on this when I have a moment and hopefully post my new answer soon.

a=g(sin)(theta)-(mu)g(cos)(theta)
a+(mu)g(cos)(theta)=g(sin)(theta)
mu=g(sin)(theta)-a+g(cos)(theta)
mu=(9.8sin12)-0.4+(9.8cos2)
mu=2.04-9.99
mu=7.95

Looks closer, but still wrong! I've done problems like these before, I'm not sure what's messing me up.

Medgirl314 said:
a=g(sin)(theta)-(mu)g(cos)(theta)
a+(mu)g(cos)(theta)=g(sin)(theta)
mu=g(sin)(theta)-a+g(cos)(theta)
Check that last step.

What is the force of friction? Ff = μmgcosΘ
What is the slope directional component of the pull of gravity? Fs = mgsinΘ
The resulting force Fr = Fs + Ff , where Ff is directed the other way hence its value is negative.
If you write it out with the given data then:
ma = mgsinΘ - μmgcosΘ -> ma - mgsinΘ = -μmgcosΘ
gsinΘ - a = μgcosΘ -> μ = (gsinΘ - a)/gcosΘ

I find it to be roughly around 0.17 , quite realistic since the piano has wheels.

haruspex said:
Check that last step.

Do the g's cancel out? If so, I think I could do this:

a+m(cos)(theta)=sin(theta)
m=sin(theta)-a+cos(theta)

But plugging it in and solving yields 0.2-1.38, which is better, but still too large!

Edit: What if I divide?
m=sin(theta)/a+cos(theta)
m=.208/1.28
m=.16 !

That looks right! Could someone refresh my memory on how I know whether to subtract or divide?

Last edited:
Medgirl314 said:
Could someone refresh my memory on how I know whether to subtract or divide?
The rule is that you must do the same to both sides of the equation.
From
a+(mu)g(cos)(theta)=g(sin)(theta)
you subtracted a from each side to give:
(mu)g(cos)(theta)=g(sin)(theta)-a
But then to arrive at
mu=g(sin)(theta)-a+g(cos)(theta)
you divided the left by g(cos)(theta) but added it to the right.

1 person
Ah, okay. I remembered the rule, but for some reason I didn't see that I was going against it. Is my answer correct? Thanks so much for your help!

Medgirl314 said:
Ah, okay. I remembered the rule, but for some reason I didn't see that I was going against it. Is my answer correct? Thanks so much for your help!
No, this is still wrong:
m=sin(theta)/a+cos(theta)
Take (mu)g(cos)(theta)=g(sin)(theta)-a and divide both sides by the same thing to leave just mu on the left.

(mu)g(cos)(theta)=g(sin)(theta)-a
(mu)g(cos)(theta)/g(cos)(theta)=g(sin)(theta)-a/g(cos)(theta)

Thanks! Like above?

Medgirl314 said:
(mu)g(cos)(theta)=g(sin)(theta)-a
(mu)g(cos)(theta)/g(cos)(theta)=g(sin)(theta)-a/g(cos)(theta)

Thanks! Like above?
Yes, if you put the parentheses in correctly. What you have written is g sin(θ)-(a/g)cos(θ), which is wrong. (And why do you put sin and cos in parentheses?)

Oops, forgot to reply! I put sin and cos in parentheses because I hoped it would make it easier to read. How's this?

(mugcostheta)/(gcostheta)=(gsintheta-a)/(gcostheta)

You don't have to use theta for the angle if it is uncomfortable for you to type in the smybol for it. You could simply say "theta" = B, for example, and write it as
mu gcos(B) / gcos(B) = (g sin(B) - a) / g cos(B)

The value of a function is taken from the product, you can add parenthesis to make it clear. There is no need to add parenthesis around the name of a function.

Mostly you add parenthesis based on the hierarchy of operations, but that's for another thread.

Medgirl314 said:
Oops, forgot to reply! I put sin and cos in parentheses because I hoped it would make it easier to read. How's this?

(mugcostheta)/(gcostheta)=(gsintheta-a)/(gcostheta)
No, you should put the arguments in parentheses: cos(theta), etc. For the rest, you could use spaces: mu g cos(theta), or dots or asterisks: mu.g.cos(theta), mu*g*cos(theta). Better still, select "Go Advanced" and pick out the Greek letters from the Quick Symbols panel: μg cos(θ). (That takes a little getting used to because the symbol appears in the typing area as selected text, so if you now continue typing it will replace your symbol. You just need to click to the right of the symbol to deselect it before continuing.)

mu*g cos(theta)/g*cos(theta)=g*sin(theta-a)/g*cos(theta)

Is that better?

Thanks!

## What is the Coefficient of Friction of a Piano?

The Coefficient of Friction of a piano is a measure of the amount of resistance between two surfaces in contact with each other. It is a dimensionless number that ranges from 0 to 1, with lower values indicating less friction and higher values indicating more friction.

## Why is the Coefficient of Friction important for a piano?

The Coefficient of Friction is important for a piano because it affects the playability and sound of the instrument. A higher coefficient of friction can make the keys feel heavier and more difficult to press, while a lower coefficient of friction can make the keys feel lighter and easier to press.

## How is the Coefficient of Friction measured for a piano?

The Coefficient of Friction for a piano is typically measured using a device called a tribometer. This device measures the force required to move a surface across another surface and calculates the coefficient of friction based on that force.

## What factors can affect the Coefficient of Friction of a piano?

There are several factors that can affect the Coefficient of Friction of a piano, including the type of materials used for the keys and key bed, the condition of the keys (e.g. whether they are polished or worn), and the humidity and temperature of the environment.

## How can the Coefficient of Friction be adjusted for a piano?

The Coefficient of Friction for a piano can be adjusted by changing the materials used for the keys and key bed, by polishing or replacing worn keys, and by controlling the humidity and temperature of the environment. Additionally, lubricants can be applied to reduce friction between the keys and key bed.

• Introductory Physics Homework Help
Replies
12
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
803
• Introductory Physics Homework Help
Replies
24
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
16
Views
2K