 |
 |
Friction and Tangency
Name: Kenn P.
Status: other
Age: 50s
Location: N/A
Country: N/A
Date: 2000-2001
Question:
I am the Mathematics Test Specialist for GED Testing
Service, the organization that develops the High School Equivalency
Exam. One of my Item Writers submitted a question in which she stated
that the coefficient of friction was equal to the tangent of the angle of
inclination at which one surface begins to move across another. I seem
to recall from my high school physics class (in 1963) that such a
statement was made, but perhaps that explanation is out of date. I would
appreciate any attempt to verify the accuracy of the statement.
Replies:
Tipler's Physics text (Worth, 1976) shows how this result derives from
Newton's equation. (It's the coefficient of static friction they're
talking about, of course.)
Tim Mooney
Kenn,
This statement is true for the coefficient of STATIC friction. The
coefficient of DYNAMIC friction is smaller.
Consider an object on a ramp, just starting to overcome static friction. If
we let "A" be the angle of the ramp, the force of gravity pulling the object
down the ramp is scaled by sin(A): F=mg.sin(A). The normal force between
the object and ramp, how hard they are pressed together, is scaled by
cos(A): N=mg.cos(A). As e-mail doesn't deal well with Greek symbols, we
will use "u" as the coefficient of static friction. At the point of just
overcoming friction, the force down the ramp equals the frictional force.
The frictional force is "u" times the normal force.
F = u.N
mg.sin(A) = u.mg.cos(A)
divide out mg
sin(A) = u.cos(A)
sin(A) /
cos(A) = tan(A)
tan(A) = u
Dr. Ken Mellendorf
Illinois Central College
Click here to return to the Physics Archives
| |
Update: June 2012
|
|