# With what minimum acceleration mass M be moved on frictionless surface so that m remains stick to it. The co-efficient of friction between M and m is u?

Sep 15, 2017

If $\mu$'s value is any less than $\frac{a}{g}$, m will slip -or- if a's value is any greater than $\mu \cdot g$, m will slip.

#### Explanation:

I will assume that mass m is sitting on top of mass M.

I still can not see meaning in the question as written. It would make better sense to ask for 1) minimum co-efficient of friction or 2) maximum acceleration. I will solve it both ways.

First interpretation:
What is the minimum co-efficient of friction between M and m that allows M to be moved at acceleration a, yet m remains stuck in its place on M.

The force of friction, ${F}_{f}$, between M and m is given by

${F}_{f} = \mu \cdot N = \mu \cdot m \cdot g$
where N is the normal force of m on the surface of M and g is the acceleration due to gravity (therefore mg is m's weight).

If m is to be remain in its place on M while M is moving with acceleration a, then m must move with the same acceleration that M moves. Therefore, m must be given a force F according to Newton's 2nd Law:

$F = m \cdot a$

The only source for the force F is the friction between M and m. Therefore the force F can not be more than the value of F_f. Therefore, at the limit,

${F}_{f} = m \cdot a$
$\mu \cdot m \cdot g = m \cdot a$
Cancelling m and solving for $\mu$
$\mu \cdot \cancel{m} \cdot g = \cancel{m} \cdot a$
$\mu = \frac{a}{g}$

If $\mu$'s value is any less, m will slip.

Second interpretation:
What is the maximum acceleration that allows M to be moved at acceleration a, yet m remains stuck in its place on M.

Note: the beginning of the work in this second interpretation will be identical to the beginning of what is above for the first interpretation. It is repeated here for completeness.

The force of friction, ${F}_{f}$, between M and m is given by

${F}_{f} = \mu \cdot N = \mu \cdot m \cdot g$
where N is the normal force of m on the surface of M and g is the acceleration due to gravity (therefore mg is m's weight).

If m is to be remain in its place on M while M is moving with acceleration a, then m must move with the same acceleration that M moves. Therefore, m must be given a force F according to Newton's 2nd Law:

$F = m \cdot a$

The only source for the force F is the friction between M and m. Therefore the force F can not be more than the value of F_f. Therefore, at the limit,

${F}_{f} = m \cdot a$
$\mu \cdot m \cdot g = m \cdot a$
Cancelling m and solving for a
$a = \mu \cdot g$

If a's value is any greater, m will slip.

I hope this helps,
Steve