The wooden crate from the last lesson is sliding across its concrete floor again. For that pair, the coefficient of friction is 0.40.
Suppose the normal force on the crate today is 150 N. How large is the kinetic friction on it?
The friction should be 0.40 of 150 N. There is an equation that says exactly that.
The fraction rule, written as arithmetic: kinetic friction = coefficient × normal force. With the numbers: 0.40 × 150 = 60. The kinetic friction is 60 N.
In symbols, the kinetic friction law is:
f_k = μ_k N
μ_k is the pair's coefficient for sliding, and f_k is the sliding friction it gives.
One care point: in this equation, the symbol N stands for the normal force on the object. It is not the unit newton.
The same letter does two jobs in physics. In this equation, the letter is the force.
The equation is a recipe: multiply the pair's μ_k by the normal force, and the product is the kinetic friction force, in newtons.
A plastic storage tub slides across a tile floor. μ_k = 0.30 for the pair, and the normal force on the tub is 180 N. Calculate the kinetic friction on the tub.
Write down the values in the question:
μ_k = 0.30
N = 180 N (the normal force)
f_k = ?
Write down the equation:
f_k = μ_k N
Substitute in the values:
f_k = 0.30 × 180
Simplify:
f_k = 54 N
The kinetic friction on the tub is 54 N.
A crate slides across a shop floor. μ_k = 0.25 for the pair, and the normal force on the crate is 120 N. Enter the kinetic friction on the crate in N.
Tilda reads f_k = μ_k N and says: 'The N in this equation is the normal force on the sliding object.' Is she correct?
Rowan reads f_k = μ_k N and asks whether the letter N at the end means the answer is already in newtons. What is the N in the equation?
The N in the equation is a symbol for a force: the normal force on the sliding object.
You substitute that force's size, in newtons, in its place.
The unit newton is also written N, but after a number, as in 54 N. In the equation, the letter is the force.
The equation's N is the normal force on the object, not the unit newton.
Imre reads f_k = μ_k N and says: 'The N at the end is the unit newton, so the equation already gives the answer in newtons.' Is he correct?
A rubber block slides across a lab bench. μ_k = 0.60 for the pair, and the normal force on the block is 85 N. Enter the kinetic friction on the block in N.
Which statement reads f_k = μ_k N correctly?
A loaded sled slides across packed snow. μ_k = 0.15 for the pair, and the normal force on the sled is 440 N. Enter the kinetic friction on the sled in N.