# Intuitive Understanding of Mass, Acceleration, Friction Force and Coefficient of Friction?

When calculating maximum acceleration, I know that mass cancels out to get:

$$\mu_s = a/g$$ Is there an intuitive approach to this?

Here's a problem that involves this:

If half of the weight of a small $1.00 \times 10^3$ kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete?

• Please give us more to help us understand the question. I suspect that you are addressing the odd fact that the $m$ in the force of gravity $mg$ cancels the $m$ in Newton's Law. But I can't really tell what you are asking. In particular, what you mean by "maximum acceleration (before an object moves)". If there's any acceleration, the object is already moving. Also, what is $u$? – garyp Nov 14 '16 at 2:52
• @garyp updated. – Imagine Dragons Nov 14 '16 at 3:09
• $\uparrow$ mew=mu? – Qmechanic Nov 14 '16 at 7:15
• @ImagineDragons What is mew-coeff. of friction? – Prashant Singh Nov 14 '16 at 7:35
• Thanks for the clarification. I've edited your question to make it more clear. If I've made any changes that you don't like, go ahead and change them. In the future, please be careful to include enough information so that someone who is not sitting next to you can understand the question. Also, please set all equations in MathJax. Here’s a MathJax tutorial. Without doing those two things, most people will ignore the question because it can't be understood. – garyp Nov 14 '16 at 16:01

Is there an intuitive explanation? I don't think so. You have lost some understanding by cancelling $m$ :
$$\mu_s=F/N=ma/mg=a/g$$ $\mu$ is defined to be the ratio of 2 forces. It is an experimental fact that friction and normal reaction are proportional for a wide range of values for $N$.
All the expression $\mu=a/g$ says is that $\mu$ scales proportionaly with $a$ (because $g$ is a constant).
But this is not any $a$; it is the acceleration $a$ which friction would cause alone! Only in this case, can this expression be used.
And since this $a$ scales proportionally with friction (this is Newton's 2nd law), and since $\mu$ also scales proportionally with friction (from the formula $f=\mu N$), then of course $\mu$ scales proportionally with this $a$.
But again watch out! This is not any $a$. This expression cannot be used in any cases with whatever the $a$ is.