How Do You Calculate the Frictional and Coupling Forces for a Skidding Truck?

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Discussion Overview

The discussion revolves around calculating the frictional and coupling forces acting on a truck that is skidding to a stop. Participants explore the application of physics principles to determine these forces based on given parameters such as initial velocity, distance skidded, and mass.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents the problem of a truck skidding and suggests using the equation v² = v₀² + 2a(s - s₀) to find deceleration.
  • Another participant agrees that finding deceleration is a valid approach and recommends drawing free body diagrams for the truck and trailer to apply F = ma.
  • A participant calculates forces FB and FT for the truck and trailer, respectively, and attempts to find the coupling force FC by subtracting FB from FT.
  • There is uncertainty regarding the coefficient of friction µ, with one participant suggesting an assumption of µk = 0.8 for dry concrete.
  • Another participant argues that µ is irrelevant for determining the forces and points out that the F = ma equations do not account for the coupling force between the truck and trailer.
  • Further clarification is provided that the trailer experiences only one horizontal force, while the truck has two forces acting on it, assuming the trailer wheels are not skidding.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of the coefficient of friction and the inclusion of coupling forces in their calculations. The discussion remains unresolved regarding the correct approach to incorporate these forces into the equations.

Contextual Notes

There are limitations in the assumptions made about the forces acting on the truck and trailer, particularly regarding the skidding of the trailer wheels and the moment of inertia. The discussion also reflects uncertainty in applying the correct values for the coefficient of friction.

winblowzxp
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1.The 2-Mg truck is traveling at 15 m/s when the brakes
on all its wheels are applied, causing it to skid for a distance of
10 m before coming to rest. Determine the constant horizontal
force developed in the coupling C, and the frictional force
developed between the tires of the truck and the road during
this time. The total mass of the boat and trailer is 1 Mg.


v0 = 15 m/s
s = 10 m
Fk = µkN

2. I'm not exactly sure how/where to start. I also have no idea how to get the frictional force.



The Attempt at a Solution


I figure that since I know v0 and vf as well as s0 and sf that I can get deceleration.

I'll use v2 = v02 + 2a(s-s0). If I solve for a, maybe that will help get the frictional force?
 
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welcome to pf!

hi winblowzxp! welcome to pf! :wink:
winblowzxp said:
I'll use v2 = v02 + 2a(s-s0). If I solve for a, maybe that will help get the frictional force?

yes, that will give you the deceleration …

then draw two free body diagrams, one for the truck and one for the trailer, and do F = ma for each …

what do you get? :smile:
 
I get:

FB = -11.25 kN
FT = -22.5 kN

I figure that to get the force in the coupling (C), that I need to subtract FB from FT which would give me FC = -11.25 kN



I know that fk = µN and that N in this case is -W. I'm not sure how to get the value for µ.
[PLAIN]http://www.fission-systems.com/images/physics/fma.png


Edit: I think I'll assume that it's on dry concrete and use µk = 0.8.
 
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i'm really not following that :confused:

µ is irrelevant, the forces can be found from the masses and the acceleration

your F = ma equations don't seem to include the force in the coupling (between the truck and the trailer) :redface:
 
your F = ma equations don't seem to include the force in the coupling (between the truck and the trailer)

I'm not exactly sure how to do that.
 
the trailer has only one horizontal force on it, the truck has two :wink:

(i'm assuming that the trailer wheels are not skidding, and that their moment of inertia is sufficiently small that you can ignore the friction)
 

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