How to Calculate the Coefficient of Friction
The coefficient of friction comes in two forms: static and kinetic. The coefficient of static friction is a force between two surfaces that are in contact but are at rest. The coefficient of kinetic friction tells us how rough or smooth the frictional force is between two moving surfaces. To calculate either, it is necessary to know the equation for the frictional forces and Newton's Second Law.
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Instructions
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Instructions
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1
Learn and understand the equation used to find the coefficient of static friction. It is a friction of rest in that it opposes any movement, and its maximum value is f(s) = u(s)N, where u(s) is the coefficient of static friction, and N is the normal force. The coefficient of static friction, therefore, is f(s)/N, a dimensionless number. A body that overcomes the static frictional force begins to move, and this force then changes to f(k).
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2
Learn and understand the equation used to find the coefficient of kinetic friction, which is weaker than the coefficient of static friction. The coefficient of kinetic friction opposes the movement of surfaces where one or more is in motion. Its magnitude is f(k) = u(k)N, where u(k) is the coefficient of kinetic friction and N is the normal force. The coefficient of kinetic friction, therefore, is f(k)/N. Like u(s), it is a dimensionless number.
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Use the equation given in step one to find the coefficient if you have a problem where you are given f(k) or f(s) or both as well as N. Plug in the values and solve directly for u(k) or u(s). If you are not given the frictional and normal forces, proceed to Step 4.
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4
Calculate f(s), f(k) and N by using Newton's Second Law, F = ma. Find the sum of all of the forces acting upon the moving object. Remember that for no acceleration, the net forces will be equal to zero. Remember also that f(s) and f(k) oppose movement and so will be opposite in sign to the force making the object move.
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Draw a free-body diagram to find the horizontal and vertical components of the forces acting upon the object in Step 3. A standard example is a box being pushed or pulled.
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Practice Steps 4 and 5 by studying the diagram to the left and writing out the appropriate equation. Shown is a box initially at rest that must be pulled with a force, F, before it begins to move. Before movement, f(s) is at a maximum.
The sum of all the forces: F + N + f(s) + mg = 0.
The sum of all the horizontal forces: F - f(s) = 0 (there is no acceleration).
The sum of all of the vertical forces: N - mg = 0 (the object is not moving up or down).
Solving these simultaneously, u(s) = F/mg. If the pulling force is 250 N and the weight of the box is 600 N, u(s) = 0.416, where u(s) is at a maximum. -
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Practice Steps 4 and 5 by studying the diagram to the left, and writing out the appropriate equation. The box is now moving by being pulled with a force, F, that is less than the one in Step 6.
The sum of all the forces: F + N + f(k) + mg = 0.
The sum of all the horizontal forces: F - f(k) = 0 (there is no acceleration).
The sum of all of the vertical forces: N - mg = 0 (the object is not moving up or down).
Solving these simultaneously, u(k) = F/mg. If the pulling force is 220 N and the weight of the box is 600 N, u(s) = 0.37.
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Tips & Warnings
The values of u(s) and u(k) depend on the smoothness of the surface. Smooth surfaces have lower values than rougher ones. These numbers are approximate, depending on whether the surfaces are clean or wet or sanded, for example.
References
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- Photo Credit Sasa Stefanovic
