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Question
A piano weighing 4880 n slides down an incline plane at a constant velocity. The plane is 20.0 meters longs. Its height is 2.50 meters. Calculate:
A. The component of the piano's weight parallel to the incline.
B. the component of the piano's weight perpendicular to the incline
C. the coefficient of the sliding friction.

Answer
As the piano slides down the incline at a constant speed the external forces acting on the piano will be:
Fg - the force of gravity, straight down
Ff - the force of friction, parallel to and directed up the incline
Fn - the normal force exerted on the piano bevy the incline, directed perpendicular to and away fom the incline
Since the gravitational force is straight down you will need to break that force into components parallel to Fpar and perpendicular to Fper the surface of the incline. The component ofthe force parallel to the incline will be equal o the force of gravity Fg multiplied by the sine of the angle of the incline:
Fpar=Fg*sin(alpha)
The angle alpha is equal to the inverse sine of the height h of the incline divided by the length L of the incline:
alpha-=arcsin(h/L)=arcsin(2.5/20)=7.18 deg
Therefore, the component of the gravitational force parallel to the incline will be:
Fpar=Fg*sin(alpha)=4880*sin(7.18)=610N
The component of the gravitational force perpendicular to the incline will be equal to the gravitational force Fg multiplied by the cosine of the angle of the incline:
Fper=Fg*cos(alpha)=4880*cos(7.18)=4842N
Because the piano is moving down the incline at a constant speed opposite forces are equal. Therefore, the force of friction Ff directed up the incline must be equal to the gravitational force component Fpar directed down the incline:
Ff=Fpar=Fg*sin(alpha)
While the normal force Fn must be equal to the gravitational force component perpendicular Fper to the incline:
Fn=Fper=Fg*cos(alpha)
Since the force of friction Ff is equal to the product of the normal force Fn and the coefficient of sliding friction mu in this case:
Ff=Fn*mu
This becomes:
Fg*sin(alpha)=Fg*cos(alpha)*mu
Solving for the coefficient of sliding friction mu:
mu=Ff/Fn=Fg*sin(alpha)/Fg*cos(alpha)=sin(alpha)/cos(alpha)=tan(alpha)=tan(7.18)=0.126

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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