Train pulling caboose - forces and friction

[BlueSkyy]

Homework Statement

A 2.7 kg toy locomotive is pulling a 1.2 kg caboose. The frictional force of the track on the caboose is 0.48 N backward along the track. If the train is accelerating forward at 2.7 m/s2, what is the magnitude of the force exerted by the locomotive on the caboose?

Homework Equations

F=ma and f=uN (the formula relating friction to normal force)

The Attempt at a Solution

I've found the normal forces and weights and drawn my FBD's (free body diagrams) and realize that the tension is the only force acting on the caboose other than friction...
Am I supposed to add the masses of the objects, and find the force using the acceleration? Do I subtract friction?

I know:
2.7 kg X 2.7 m/s2 = 7.29 N (mass of train times accel.)
1.2 kg X 2.7 m/s2 = 3.24 N (mass of caboose times accel.)
(1.2 kg + 2.7 kg) X 2.7 m/s2 = 10.53 N (both masses combined times accel.)

i also assume friction needs to be subtracted since it acts in the negative direction...but i don't know where to go from here...

 

[Dick]

You know one force on the caboose, F_friction=0.48N backwards. You know the acceleration of the caboose because it's the same as the train in a forward direction. As before, in F=ma, F is the total of all the forces. So F=F_friction+F_locomotive. Can you find F_locomotive? You don't need to worry about mu and normal forces, you are already given the frictional force. You don't have to compute it.

 

[BlueSkyy]

I've tried entering every answer I can think of and they all turn out wrong...
When using F=ma regarding the train, you would take 2.7 kg X 2.7 m/s2 = 7.29 N, correct? Then adding the force of friction (.48 N) you would get 7.77 N? Which is incorrect, as is 6.81 N (what would have happened had you subtracted .48 N)

 

[Dick]

The force of the locomotive and the force of friction are in opposite directions, right. So you subtract. But the mass in F=ma is the mass of the caboose, right? That's the object you are calculating forces and accelerations for. You don't need to to know the mass of the locomotive at all. It's just there to distract you.

 

[BlueSkyy]

I've also tried 3.24 N (mass of caboose times accel.) plus and minus the .48 N, with no success.

 

[Dick]

(-0.48N+F_loco)=(1.2kg)*2.7m/sec^2. If you got 3.72N, then you are correct. And I don't know what else to say.

 

[BlueSkyy]

Yup, that is what I got...must be an error in the assignment.

Thank you for all your time and help!
God bless~

 

[Dick]

You have two 2.7 numbers in the problem. Are you sure they are both correct?