Static Friction but no mass given! Help!

1. Sep 20, 2011

cassienoelle

1. The problem statement, all variables and given/known data

The coefficent of static friction between the floor of a truck and a box resting on it is 0.36. The truck is traveling at 70.0 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?

Professor gave us these clues:
It is the static frictional force exerted on the box by the truck that causes the box to accelerate. The static frictional force is:
fs≤μs*N
Since we're are interested in stopping in the least amount of time possible, the frictional force must be equal to the maximum possible static frictional force (which results in the largest acceleration):
fs=μs*N

2. Relevant equations

so if fs=Us*N

3. The attempt at a solution
where the heck can i find N and fs? If I have no mass given i can't find N?
I've figured out that the speed is 19.44m/s :)

2. Sep 20, 2011

PeterO

The frictional force is Us*N as you say. That means Us*m*g.
Once you had the frictional force, you would divide by m to get the resulting acceleration so you could calculate the stopping distance. That means the mass would cancel out, croceed through the problem using m for mass and it will disappear at a convenient time.

EDIT: In other words, you are not really after the actual Frictional force, just the affects the friction force will have.

3. Sep 20, 2011

cassienoelle

So...
fs=.36*m*9.81
So...
fs=3.5316m
I don't understand how the m's would cancel out?

4. Sep 20, 2011

Aggression200

I have a similar question (It's the same form, just different numbers).

So, you would have....

f$_{s}$=$\mu$N

You need the least distance needed to stop, so you would start by finding the maximum acceleration, correct?

I will denote m$_{t}$ as the truck's mass and a$_{t}$ as the acceleration of the truck.

So, f$_{s}$=$\mu$m$_{t}$g = m$_{t}$a$_{t}$

So, the masses cancel out, leaving....

$\mu$g = a$_{t}$

So, a$_{t}$ is .36(9.8)....

a$_{t}$ = 3.53 m/s^2 Am I on the right track?

5. Sep 20, 2011

PeterO

SO you have an expression for fs. What are you going to use it to calculate?

6. Sep 20, 2011

PeterO

Looks correct.

7. Sep 20, 2011

Aggression200

Thanks, PeterO.

Does this method from here on seem right?

Since I have a, V$_{i}$, and V$_{f}$, I can now solve for $\Delta$x using kinematics....

V$_{f}$$^{2}$=V$_{i}$$^{2}$+2a$\Delta$x

V$_{f}$=0 m/s
V$_{i}$= 19.44 m/s
a = -3.53 m/s^2

So....

$\Delta$x = $\frac{-(19.44)^2}{-2(3.53)}$

So, $\Delta$x = 53.5 m ?

Last edited: Sep 20, 2011
8. Sep 20, 2011

PeterO

Assuming you arithmetic was correct that should be true.
rough check is: average speed approx 10m/s
time to stop approx 5 seconds
Distance covered approx 50m ????

You are out by a factor of 2 ??

I bolded and coloured a 2 in one of your formulas. have you perhaps overlooked it during your calculations?

9. Sep 20, 2011

Aggression200

Yes, I caught it right before you posted. Sorry. I had it written down on my paper, but typed it in wrong.

Thanks for all your help, PeterO.

10. Sep 20, 2011

cassienoelle

nevermind! THANK YOU THANK YOU THANK YOU!

11. Sep 20, 2011

Aggression200

You're very welcome. Sometimes it just helps to see other people work it out.

12. Sep 20, 2011

cassienoelle

The class I'm taking is online, and he hasn't showed us how to work these out very well. So it's extremely frustrating. Thank you for understanding.

13. Sep 20, 2011

cassienoelle

Nevermind i got it all by myself :d

14. Sep 20, 2011

Aggression200

I understand completely. Teachers in the classroom who don't explain it very well make it extremely frustrating, much less online.