Homework Help: Frictional force by tires

1. Jul 21, 2011

kraigandrews

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

An SUV pulls a trailer carrying a boat. The mass of the SUV is 2460 kg and the mass of the boat and trailer is 490 kg. The SUV starts from rest (velocity = 0) and accelerates for 10 seconds to final velocity 12 m/s; i.e., a = 1.2 m/s2.

What is the frictional force exerted by the road on the tires of the SUV?

2. Relevant equations

F=ma

3. The attempt at a solution

so I used
.5mv^2=$\mu$mgd

d=.5at^2
d=60m

thus $\mu$=0.1223

so then Ff on the road by the tires =M_suv*0.1223=300.92 N, but this is not correct, where am I going wrong this seems so simple

2. Jul 21, 2011

Pi-Bond

Why are you equating the kinetic energy and energy dissipated by friction? No kinetic energy is being dissipated by friction here. Use Newton's second law.

3. Jul 21, 2011

kraigandrews

ok yeah I thought that was wrong, so then the net force on the SUV is 2950N, but I am confused as to where to go from here.

4. Jul 21, 2011

Pi-Bond

Is this all the information you have? You can express the force on the SUV in terms of the frictional force and a forward force.

5. Jul 21, 2011

kraigandrews

This is all the info I have yes. So could you nudge me in the right direction at least, because at the surface this seems simple so I think I'm over looking something.

6. Jul 21, 2011

Pi-Bond

When the SUV obtains it's final velocity, all forces on it must be balanced. So you need to find the forces on the SUV.

7. Jul 21, 2011

kraigandrews

so the forces needed here are just the forward force and frictional, so F=ma however how do you find the Forces with the info given, I would only think the net force can be found

8. Jul 21, 2011

Pi-Bond

True, you need the forward force. I might be missing something here...let me think.

9. Jul 21, 2011

Pi-Bond

Do you have to find the frictional force or the coefficient of friction?

EDIT: It seems your original method is correct. From conservation of energy-

0=(Kinetic Energy)+(Work due to friction)

I think you forgot a g when calculating the frictional force. Instead of taking that long route you could just have divided the kinetic energy by the distance. F=ma isn't really applicable since you lack knowledge of all forces.

Last edited: Jul 21, 2011