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October, 2006
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Cleophas "Mike" McAlpin...Executive Director/Chief Tutor
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October, 2006
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The Tutor Speaks: Shalay, this is not a hard problem, although it is a 10th-12th grade level one. In preparation, we will review the basic trigonometric functions dealing with the sine, cosine and tangent of a right triangle. Afterwards, we will apply logic to the problem of solving for "a" ,the acceleration of a body that has an unbalanced force acting upon it! The body being the box shown, of course. We will revisit this problem after 2 weeks of intense tutoring! Corrections? send E-mails to cleophas9@sbcglobal.net
Box on an Inclined Plane with
Friction: An application of
Newton's 2nd Law of Motion
A mass sliding down an inclined plane
with friction experiences three forces. One is the force of gravity (weight),
which pulls down on the box in the vertical direction.
The second force is the opposing normal
force, "N" of the plane, which pushes up on the box in a direction that is
perpendicular to the surface of the inclined plane ( MGCOSǾ
). The third force is friction, (F=µN,
where N = MGCOSǾ), F=
µ (MGCOSǾ)
which opposes the motion and is parallel to the plane. The
acceleration of the mass sliding along a
plane (inclined at an angle of (Ǿ)
is determined by the net force, which consists of the component of the weight
vector down (MGSINǾ
) the plane minus the magnitude of the friction force,
µ (MGCOSǾ) .
MGSINǾ - µ (MGCOSǾ) = MA= F
Actually, the mass (M) is immaterial, seeing that it can be eliminated from the equation!
Then we have: GSINǾ - µ (GCOSǾ) = A, which is the acceleration of the body down the plane.
We can take out the "G" on the left side of the equation to yield:
G(SINǾ - µ COSǾ ) = A
And so, Mr. Isaac Newton, we have arrived at your "master formula" for sliding bodies: utilizing your second law of motion, F=MA
Several observations stand out when we examine your formula. For example, if the body is dropped vertically, the cosine of 90° = 0, and so there is no "µ COSǾ"; just the GSINǾ: Since the Sine of 90° = 1, then the acceleration becomes "G" or 32.2 Ft/sec2 ( Lesson Review 11 B )
The first step is to draw a *free-body
diagram* showing those forces that act on the box.
The box is sliding down the plane to the
right or down the plane, so the friction force opposing the motion must act
upward along the plane to the left. In the y-direction (perpendicular to
the plane) the component of the box's weight is downward and the normal force is
upward. Because the box does not fly off into the air or sink down through
the inclined plane, these forces must sum to zero....Example: Let the angle to
the horizontal be 30°:
The weight of the box is 100 pounds (who cares about the
weight?); the coefficient of friction,"µ"
,mu, is equal to 0.20. What is the acceleration of the box as it slides down the
plane?
Solution: G(SINǾ - µ COSǾ ) = A
32.2 (sin 30° - 0.20 Cos 30°) = A
(32.2) ( .5 - (0.20)(0.866) = A
(32.2) ( 0.50(0.172) = A
A = 2.77Ft/sec2 Not very fast at 30°! Of course, the more we make the angle, the faster the box goes; at 89°, the box is really flying down the plane! Why? What if the coefficient of friction was equal to 0.95? 0.001?
Under Construction!
Always, Humprh!