I'm aware that this is extremely basic but I've forgotten how to get the coefficient of friction when given a forward force and acceleration.

Can someone describe to me the algorithm for solving the following example question?

Mass = 50kg

Forward Force Acting on Mass = 100 N

Acceleration of Mass = 0.1m/s^2

Coefficient of friction?

EDIT: Am I right in thinking it is 0.2?

  • $\begingroup$ Have you revisted an introductory physics textbook, or looked online at recourses like hyperphysics? $\endgroup$ – Cicero Jun 10 '15 at 13:40
  • $\begingroup$ I have but I didn't see anything that regarded getting the coefficient of friction when acceleration is included. I am having somewhat of a mental-blank at the moment and just want to make sure I know the steps needed as I have a physics test later today. EDIT: My mental fog is clearing and I am beginning to realize ridiculously easy this is. I'd still like an answer to assure I am correct but I'm fairly sure I can figure this out on my own $\endgroup$ – RJGordon Jun 10 '15 at 13:42
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    $\begingroup$ There are two forces, the forward force and friction. The difference of these forces is the net force equal to the product of mass and acceleration. From this you get the friction force. From this divide then normal force, mg to get them friction force. $\endgroup$ – Cicero Jun 10 '15 at 13:53
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    $\begingroup$ It is 0.2 correct? $\endgroup$ – RJGordon Jun 10 '15 at 13:54
  • $\begingroup$ Really 0.1938...., but rounding to two decimals or using g=10, you get 0.19 $\endgroup$ – Cicero Jun 10 '15 at 13:59

Do a free body diagram and you will find for a horizontal plane that

$$ F - \mu m g = m a $$

$$ (100) - \mu (50) (9.80665) = (50) (0.1) $$

$$\boxed{ \mu = \frac{(100)-(0.1)(50)}{(50)(9.80665)} = 0.1937\ldots }$$

  • $\begingroup$ Ah yes. Immediately substitute values into the starting equation and avoid the algebra necessary to separate the unknown variable. THAT technique is why most U.S. high school students are practically algebraically illiterate. $\endgroup$ – David White Feb 24 '19 at 21:26
  • $\begingroup$ @DavidWhite - I actually I didn't, since the final expression was already solved for $\mu$ and the substitution was the last step. The difference is if I carry the symbols vs. their value around. $\endgroup$ – John Alexiou Feb 25 '19 at 0:05

Newton's 2nd law in the x-direction. There are only two forces that together equal the $ma$, and since you know one of them, it is straight forward from here.


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