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Coefficient of Friction of object sliding down ramp.. need help 
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#1
Feb810, 08:52 PM

P: 16

1. The problem statement, all variables and given/known data
An 10.0kg object slides down an inclined plane that is 10.0m in length. THe height of the incline is 6.0m and the speed of the object at the bottom is 9.0 m/s. What is the coefficient of friction of this incline? So far I found: Energy lost due to friction= 183.6 J Force of friction= 18 N Can someone show me how to find the coefficient of friction ASAP please? This is an advance question in my class for bonus marks & I NEED those bonus marks. 


#2
Feb810, 08:56 PM

P: 16

1. The problem statement, all variables and given/known data
An 10.0kg object slides down an inclined plane that is 10.0m in length. THe height of the incline is 6.0m and the speed of the object at the bottom is 9.0 m/s. What is the coefficient of friction of this incline? So far I found: Energy lost due to friction= 183.6 J Force of friction= 18 N Can someone show me how to find the coefficient of friction ASAP please? This is an advance question in my class for bonus marks & I NEED those bonus marks. 


#3
Feb810, 08:59 PM

P: 16

oh yeah i also found the angle at the bottom is 36.869... degrees if that helps.



#4
Feb810, 09:00 PM

P: 16

Coefficient of Friction of object sliding down ramp.. need help
oh yeah i also found the angle at the bottom is 36.869... degrees if that helps



#5
Feb810, 09:02 PM

P: 344

The formula relating friction force and coefficient of friction is: f = u*Fn, where f is the friction force, u is mu(coefficient of friction), and Fn is the normal force.



#6
Feb810, 09:04 PM

P: 16

how do I find the normal force?
I just started physics 30 & totally forget everything... :( 


#7
Feb810, 09:05 PM

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P: 6,035

You have the friction force, so now what formula relates the friction force with the coeff. of friction??



#8
Feb810, 09:06 PM

P: 16

u= F/ Fn but i don't know what my Fn is?



#9
Feb810, 09:10 PM

P: 16

okay i did tried this:
Fn= (10.0kg)(cos(36.869...degrees))(9.81) = 78.48...N u=F/Fn =0.22935... is this correct? 


#10
Feb810, 09:14 PM

P: 344

the normal force is the force the object exerts on whatever surface it rests. In other words, its mass(i.e. m*g)



#11
Feb810, 09:14 PM

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P: 6,035

Ok, Fn is the normal force acting on the object perpendicular to the incline. Call that direction the 'y' axis, and the direction along the incline will be the 'x' axis. In the y direction, there is no acceleration, so Newton's first law applies. Identify all the forces or force components in the y direction (what are they?) and set their sum equal to 0 to solve for Fn .



#12
Feb810, 09:14 PM

P: 16

okay i did tried this:
Fn= (10.0kg)(cos(36.869...degrees))(9.81) = 78.48...N u=F/Fn =0.22935... is this correct? 


#13
Feb810, 09:19 PM

P: 16

I don't know the other forces, only the top mechanical energy & bottom mechanical energy



#14
Feb810, 09:20 PM

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PF Gold
P: 6,035




#15
Feb810, 09:21 PM

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PF Gold
P: 6,035




#16
Feb810, 09:23 PM

P: 16

i figured out the 18 N in class but I don't remember how but thanks a bunch for the help, even though u say I didn't need the hint, your hint made me think & brought back everything! =)



#17
Feb810, 09:38 PM

P: 344

Yes, that's correct.



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