Static Friction but no mass given Help

[cassienoelle]

Homework Statement

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

Homework Equations

so if fs=Us*N

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 :)

 

[PeterO]

Homework Statement

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

Homework Equations

so if fs=Us*N

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 :)

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.

 

[cassienoelle]

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

 

[Aggression200]

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

So, you would have...

f_{s}=\muN

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}=\mum_{t}g = m_{t}a_{t}

So, the masses cancel out, leaving...

\mug = a_{t}

So, a_{t} is .36(9.8)...

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

 

[PeterO]

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

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

 

[PeterO]

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

So, you would have...

f_{s}=\muN

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}=\mum_{t}g = m_{t}a_{t}

So, the masses cancel out, leaving...

\mug = a_{t}

So, a_{t} is .36(9.8)...

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

Looks correct.

 

[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 \Deltax using kinematics...

V_{f}^{2}=V_{i}^{2}+2a\Deltax

V_{f}=0 m/s
V_{i}= 19.44 m/s
a = -3.53 m/s^2

So...

\Deltax = \frac{-(19.44)^2}{-2(3.53)}

So, \Deltax = 53.5 m ?

 

[PeterO]

Thanks, PeterO.

Does this method from here on seem right?

Since I have a, V_{i}, and V_{f}, I can now solve for \Deltax using kinematics...

V_{f}^{2}=V_{i}^{2}+2a\Deltax

V_{f}=0 m/s
V_{i}= 19.44 m/s
a = -3.53 m/s^2

So...

\Deltax = \frac{-(19.44)^2}{-3.53}

So, \Deltax = 107 m ?

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?

 

[Aggression200]

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?

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.

 

[cassienoelle]

nevermind! THANK YOU THANK YOU THANK YOU!

 

[Aggression200]

nevermind! THANK YOU THANK YOU THANK YOU!

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

 

[cassienoelle]

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

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.

 

[cassienoelle]

Nevermind i got it all by myself :d

 

[Aggression200]

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.

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