Dynamics, Coefficient of Friction

[CursedAntagonis]

Homework Statement

The coefficients of friction between the load and the flat-bed trailer shown are $$\mu$$s = 0.40 and $$\mu$$k = 0.30. Knowing that the speed of the rig is 45 mi/h, determine the shortest distance in which the rig can be brough to a stop if the load is not to shift.

The drawing shows a box sitting a top of a trailer bed with a distance of 10' between the box and the cabin.

Homework Equations

The Attempt at a Solution

I am not really sure how to begin this problem. I am thinking using the coefficient of friction I can find the acceleration of the box and use that to find the distance needed for the truck to stop, but I am not sure.

So far I have used the $$\mu$$k to find the acceleration of the box if it was to move, and was able to find the velocity of the box if it were to move.

 

[PhanthomJay]

I am thinking using the coefficient of friction I can find the acceleration of the box and use that to find the distance needed for the truck to stop, but I am not sure.

yes, but which coefficient would you use, and why?

So far I have used the $$\mu$$k to find the acceleration of the box if it was to move, and was able to find the velocity of the box if it were to move.

But if the load is not to shift, it's not supposed to move relative to the flatbed.

 

[CursedAntagonis]

yes, but which coefficient would you use, and why?

I am guessing that would be static friction since the box is to stand still while the truck is to slow down and stop.

So if I to use the coefficient of static friction to find the acceleration, and use that acceleration to find the distance it takes for the truck to stop that would be the answer?

It is a bit frustrating that the book does not have the answer to this question, it has the answer for both odds and evens and yet somehow this number is omitted.

 

[PhanthomJay]

I am guessing that would be static friction since the box is to stand still while the truck is to slow down and stop.

So if I to use the coefficient of static friction to find the acceleration, and use that acceleration to find the distance it takes for the truck to stop that would be the answer?

yes...but watch your units

It is a bit frustrating that the book does not have the answer to this question, it has the answer for both odds and evens and yet somehow this number is omitted.

so watch your math...

 

[CursedAntagonis]

yes...but watch your unitsso watch your math...

Thanks. When I find the solution, would you mind double checking my work? I will post my work.

 

[PhanthomJay]

Thanks. When I find the solution, would you mind double checking my work? I will post my work.

Not at all. But it won't be 'til morning. maybe someone else will check it if it's urgent...but it's Sunday, and nothing is due on Sunday...

 

[CursedAntagonis]

Not at all. But it won't be 'til morning. maybe someone else will check it if it's urgent...but it's Sunday, and nothing is due on Sunday...

So here is what I have got so far:

F=($$\mu$$s)N
N=mg

F=($$\mu$$s)(m)(g)
ma=($$\mu$$s)(m)(g)

the mass cancels out and thus
a=($$\mu$$s)(g)
a=(0.40)(32.2ft/s^2)
a=12.88 ft/s^2

Converting the 45 mi/h gives us 66 ft/s, and using the acceleration found in the equation V^2=Vo^2+2a(X-Xo):

0=(66ft/s)^2+2(12.88 ft/s^2)(-X) gives us
X=169.1 feet.

This is the shortest distance for the truck to stop.

Does that sound about right? Thanks for going over my work.

 

[PhanthomJay]

So here is what I have got so far:

F=($$\mu$$s)N
N=mg

F=($$\mu$$s)(m)(g)
ma=($$\mu$$s)(m)(g)

the mass cancels out and thus
a=($$\mu$$s)(g)
a=(0.40)(32.2ft/s^2)
a=12.88 ft/s^2

Converting the 45 mi/h gives us 66 ft/s, and using the acceleration found in the equation V^2=Vo^2+2a(X-Xo):

0=(66ft/s)^2+2(12.88 ft/s^2)(-X) gives us
X=169.1 feet.

This is the shortest distance for the truck to stop.

Does that sound about right? Thanks for going over my work.

Yes, good job; just round off your answer to 170 feet (2 significant figures).