Help needed for finding stopping distance?

  • #1
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Homework Statement


A tree is being transported on a flatbed trailer by a landscaper. If the base of the tree slides on the trailer, the tree will fall over and be damaged. If the coefficient of static friction between the trailer and the tree is 0.5, what is the minimum stopping distance of the truck, travelling at 15. 278 m/s, if it is to accelerate uniformly and do not have the tree slide forward and fall on the trailer?
Mu s= 0.5
V1= 15.278 m/s
d=?

Homework Equations


I'm not sure but i think
v2^2=v1^2+2ad
Us= Ffs/Fn
f=ma
v=d/t

The Attempt at a Solution


i don't really know what to do since i wasn't given the mass so i cant find acceleration or normal force[/B]
0.5(9.8)= 4.9
a= -4.9 m/s^2
v2^2-v1^2/2a=d
0^2-15.278^2?29-4.9)
d= 23.8 m
 

Answers and Replies

  • #2
Nathanael
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i don't really know what to do since i wasn't given the mass so i cant find acceleration or normal force
You can find the acceleration in terms of the stopping distance, right? Mass shouldn't be needed for that.

It's true you can't find the normal force, but just use an unknown "m" if you need to, maybe it will cancel out ;)
 
  • #3
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how would i find acceleration without the time?
and by using an unknown m do you mean Us= Ffs/ 9.8 x m ?
 
  • #4
Nathanael
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how would i find acceleration without the time?
The truck starts at 15.278 m/s and comes to a stop in a certain distance. The time it takes to stop is completely determined by that distance. (A smaller distance means the truck stopped faster.) You'll have to find a way to calculate the time (in terms of the stopping distance).

and by using an unknown m do you mean Us= Ffs/ 9.8 x m ?
Yes

Edit:
or you could use the kinematics equation and bypass the time like haruspex suggested
 
  • #5
haruspex
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how would i find acceleration without the time?
The first relevant equation you listed does not involve time.
and by using an unknown m do you mean Us= Ffs/ 9.8 x m ?
Yes.
 
  • #6
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Ffs=UFn
=0.5(9.8m)
=4.9m
i first found the frictional force and then i changed it to the acceleration
a= -4.9 m/s^2
then i plugged it into the first relevant equation
v2^2-v1^2/2a=d
0^2-15.278^2/(2 x -4.9)=d
d= 23.8 m
is this the correct way to do it?
 
  • #7
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also since i found Ffs= - 4.9 m
should i put it into F=ma?
4.9m= ma
a= 4.9m/m
and that way m gets cancelled out?
 
  • #8
Nathanael
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Looks good to me.

And yes the mass cancels out because, although the (maximum) static friction force is proportional to the mass, the force required to accelerate the mass along with the truck is also proportional to the mass, so the mass becomes irrelevant.
 
  • #9
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all right got it. so if this was a test question i would need to show the F=ma part? Thanks a lot for your help :)
 

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