Peter
Project Title
Friction Ramp
Author
Exploration by Peter Rapp '99 (7/28/99)

Issue:
    A 1kg box is placed on an inclined plane that leans against a wall to form a triangle of sides 3, 4, and 5 meters. The box does not begin to slide. Your job is to use statics to discover what value(s) of Mu (Coefficient of friction) will allow the box to stay at rest...


Discussion:
     If the numbers 3, 4, and 5 ring a bell, this is good. You probably learned in math class that this is the most common of the right triangles , used in everything from the Pythagorean Theorem proof to the declination of the shingles on your house's roof. The interior angles of this triangle are 90 (of course), 36.87, and 53.13. Try to memorize these numbers, they come up everywhere, especially in physics problems!

     The box sitting on the ramp is a simple case of statics. The net force on the object is zero, so the object is not moving. From this we know that the ups = the downs, and rightward forces = leftward forces. As Mr. Bourne always says, "Start with a free body diagram." (FBD for short) This helps you to visualize things better.

     Look at the FBD to the left. How can we compare ups to downs when the arrows are not mutually perpendicular to each other? Well, we can break them up into their components. The arrow labeled FG (Force of gravity or weight) can be divided into weight parallel to the surface and weight perpendicular. We do this with trig functions. This is pretty essential, so ask your physics teacher if you do not understand it. Note that we are using -10 here for gravity to simplify calculations. OK, here we go:

     The ramp is a 3/4/5 triangle, and the incline is the smallest angle, so we pick 36.87. This is the angle we use to break up the forces, like here:

     Frictional force is defined as Mu (the coefficient of friction) times the normal force (the ground pushing up). Weight parallel is mass times g (-10) times the trig function. We have just equated the uphill and downhill forces. We could have done the same for the forces perpendicular to the ramp, but that would not give us Mu.

     Well, if we write it this way we can easily make cancellations and cross out mg. (By the way, the three dots mean "therefore.")

     Hold it! We just canceled the weight from the equation. Does this mean that the mass of the box or even the strength of gravity do not matter when calculating Mu? Yes! The problem would be the same if the box were empty, or if it contained a lead weight; the same if we did the problem on the Earth, or on the moon.

     We don't quite have the answer yet however. Let's say that you did plug in the values given, and get 0.75 as Mu. You are not done yet. Realize that this is the minimum Mu. If Mu were 1.0, the box would hold too.


Resources Window
Note the Interactive Physics file sometimes produces erratic results when in a static situation because of inaccuracies inherent with a software integrator. If very tiny values oscillate around zero, they might just be zero =)
Text versions of the equation work are made with Microsoft Equation Editor and are embedded in Microsoft Word documents.
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