# Static Friction Between a Box and the Floor

BurpHa
Homework Statement:
A force of 35.0 N is required to start a 6.0-kg box moving
across a horizontal concrete floor. (a) What is the coefficient
of static friction between the box and the floor? (b) If the
35.0-N force continues, the box accelerates at 0.60 m / s ^ 2
What is the coefficient of kinetic friction?
Relevant Equations:
F_static = \mu_static * mg
I don't understand part (b)
In part (a), I need to calculate the coefficient of the static friction:

mg * \mu_static = 35

58.8 * \mu_static = 35

\mu_static = 35 / 58.8 \approx 0.6

So from part (a) I know that the force applied is equal to the static friction, meaning that the box cannot move (since these two forces cancel themselves out.)

However, part (b) says that the box moves at an acceleration of 0.60 m / s ^ 2.

How is it even possible for the box to even start to move and then accelerate?

Gold Member
Is the coefficient of kinetic friction usually more, less than, or equal to the coefficient of static friction ?

BurpHa
Is the coefficient of kinetic friction usually more, less than, or equal to the coefficient of static friction ?
Less than.

BurpHa
Is the coefficient of kinetic friction usually more, less than, or equal to the coefficient of static friction ?
I don't understand why the box could move when the applied force is equal to the static friction.

Gold Member
The question literally states that when 35N is applied the box moves.

• BurpHa
Homework Helper
I don't understand why the box could move when the applied force is equal to the static friction.
Ideally, if the applied force is being ramped up slowly, it will only be equal to static friction for an instant. Prior to that and it is less than static friction. More than that and the object moves and it cannot be static friction any longer.

Practically, things are not quite as clean as this. The dividing line between static and kinetic friction is not always sharp. Some times one gets things like chattering and squeaking. Also, no experiment is perfect.

@hmmm27 has it right. The question says that it moves. If you have to pretend that this takes 35.000001 N then go ahead and pretend that 35.000001 N was applied. It won't affect your results which are only good to two significant figures anyway.

Last edited:
• BurpHa and Lnewqban
Homework Helper
Gold Member
I don't understand why the box could move when the applied force is equal to the static friction.
In this case, the maximum static friction is smaller than 35.0 N; therefore, the calculated coefficient of static friction is in some degree bigger than the actual one.

• BurpHa
Gold Member
So from part (a) I know that the force applied is equal to the static friction, meaning that the box cannot move (since these two forces cancel themselves out.)
Actually, the problem states clearly that:
Homework Statement:: A force of 35.0 N is required to start a 6.0-kg box moving
across a horizontal concrete floor.
So the box moves at 35.0 N.

The reality is that ##F_{friction} < 35.0\ N## for the box not to move.

• BurpHa, berkeman and Lnewqban
Homework Helper
Gold Member
2022 Award
I don't understand why the box could move when the applied force is equal to the static friction.
This is an example of a threshold value (in this case of static friction), which seems to confuse a lot of students. You can look at this problem a couple of ways:

1) The maximum static friction is ##35.0 \ N##. In which case the forces are balanced and the block won't move. The question is how much force is required to make it move? The answer is anything in excess of ##35.0 \ N##, even if it only a tiny extra force. In that case, the applied force could still be approximately ##35.0 \ N## and the block would move.

2) The maximum static friction is almost ##35.0 \ N##. This is, in fact, what the problem is telling you. The coefficient of static friction must be almost ##0.6##.

In fact, we are given the data to one decimal place. So, if we take ##g = 9.81 m/s^2##, then the range for static friction is ##0.59## to ##0.60##, with an average of ##0.595##. See this table, giving the maximum error in the calculation.

 m F g mu_s 6.05​ 34.95​ 9.81​ 0.588875​ 5.95​ 35.05​ 9.81​ 0.600485​ 6.00​ 35.00​ 9.81​ 0.594631​

From that it makes sense that the coefficient of static friction is ##0.6## to one decimal place and perhaps ##0.59## or ##0.60## to two decimal places.

In any case, given that the data is valid only to a certain degree of precision, it makes sense that the maximum static friction force can be approximately ##35 \ N## and the minimal force needed to move the block is approximately ##35 \ N##.

• • MatinSAR, BurpHa, SammyS and 2 others
Homework Helper
Gold Member
• • BurpHa, Lord Jestocost and PeroK
Gold Member
If we will start arguing with number precision, we have to do it correctly:

The block starts moving with a force of ##35.0\ N##:
$$\mu = \frac{35.0\ N}{6.0\ kg \times 9.81\ m/s^2} = 0.5946$$
Since the most precise number has 2 significant figures, the best we can say is ##\mu = 0.59##.

Since we seem to have a 1-decimal precision for measuring the force, the highest force we can measure before the block starts moving is ##34.9\ N##:
$$\mu = \frac{34.9\ N}{6.0\ kg \times 9.81\ m/s^2} = 0.5929$$
Again, the most precise number has 2 significant figures, so the best we can say is ##\mu = 0.59##.

Either force yields the same value for the friction coefficient because of the lack of precision in the mass measurement.

If you want, you can even specify a range of values:
• ##35.0## means any number between ##34.95## and ##35.04## rounded off;
• ##6.0## means any number between ##5.95## and ##6.04## rounded off.
So the block starts moving somewhere between those two evaluations:
$$\frac{34.95\ N}{6.04\ kg \times 9.81\ m/s^2} \leq \mu \leq \frac{35.04\ N}{5.95\ kg \times 9.81\ m/s^2}$$
$$0.590 \leq \mu \leq 0.600$$
Note that we can have now 3 significant figures.

Note also that ##0.590 \times 6.0\ kg \times 9.81\ m/s^2 = 34.7\ N## or ##35\ N## when considering only 2 significant figures.

And ##0.600 \times 6.0\ kg \times 9.81\ m/s^2 = 35.3\ N## or still ##35\ N## when considering only 2 significant figures.

Still, we know for sure that the force to exceed the static friction is precisely ##35.0\ N## - not ##35\ N## - since it is a given measurement. And that means if you wanted to stay below the static equilibrium limit, only calculations with the ##0.59## value could insure that; with ##0.60##, it might go above ##35.0\ N##.

So - to be really safe - I would say that ##\mu_{static} = 0.59##.

• BurpHa
BurpHa
In this case, the maximum static friction is smaller than 35.0 N; therefore, the calculated coefficient of static friction is in some degree bigger than the actual one.

Homework Statement:: A force of 35.0 N is required to start a 6.0-kg box moving
across a horizontal concrete floor. (a) What is the coefficient
of static friction between the box and the floor? (b) If the
35.0-N force continues, the box accelerates at 0.60 m / s ^ 2
What is the coefficient of kinetic friction?
Relevant Equations:: F_static = \mu_static * mg

I don't understand part (b)
In part (a), I need to calculate the coefficient of the static friction:

mg * \mu_static = 35

58.8 * \mu_static = 35

\mu_static = 35 / 58.8 \approx 0.6

So from part (a) I know that the force applied is equal to the static friction, meaning that the box cannot move (since these two forces cancel themselves out.)

However, part (b) says that the box moves at an acceleration of 0.60 m / s ^ 2.

How is it even possible for the box to even start to move and then accelerate?

• 