What is the difference between frictional force and average force of friction?

I know we calculate frictional force in a typical slope problem by finding out the forces that act in opposition, most commonly $F_f = mg\sin\theta$ in most slope problems.

A problem is asking me to find average force of friction, and the method mentioned above is not working, so I assume that average force of friction is a whole different value. My peers have said it is calculated by using $W = F*d$ (given the values of $W$ and $d$) Why are these two different forces a different value? What is different about average force of friction?


1 Answer 1


A value of a force is usually an instantaneous value. For forces that don't change over time, that's fine.

If a force is changing over time (or might be changing over time), you might not have sufficient information to compute the value at any one instance in time. But you might have enough to compute the average value.

If a ball is sitting on a scale, the situation is static and you can calculate the force (the instantaneous force) for the ball on the scale.

If the ball is bouncing on the scale, you can't report "the force", since it changes as the ball interacts with the scale and then separates from the scale. But you could calculate an average force over time (which would be the same as the first force if losses are ignored).

  • $\begingroup$ ohhhh okay so the reason why I calculated average force of friction instead for that problem (it was about some car going up a slope with its engine off and an initial velocity) was because the force of friction was changing over time unlike in most problems where someone is continually applying force to an object to climb a slope? $\endgroup$ Jul 18 at 4:58
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    $\begingroup$ Without the problem specifics, that's difficult to know. In some cases, there is no difference and the problem writer just wrote "average force" when the force doesn't change at all. $\endgroup$
    – BowlOfRed
    Jul 18 at 5:01
  • $\begingroup$ ahhh ok I get it now! It probably did apply to my problem as I had calculated the friction force and it was a different value than the average friction force, thank you for clearing that up for me! $\endgroup$ Jul 18 at 5:06

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