A car of mass 1000kg tows a caravan of mass 600kg up a road which rises 1m vertically for every 20m of its length. There are constant frictional resistances of 200N and 100N to the motion of the car and to the motion of the caravan respectively. The combination has an acceleration of 1.2ms^-2 with the engine exerting a constant driving force. Find: a) the driving force, b) the tension in the tow-bar.
(g=10ms^-2)

Question

A car of mass 1000kg tows a caravan of mass 600kg up a road which rises 1m vertically for every 20m of its length. There are constant frictional resistances of 200N and 100N to the motion of the car and to the motion of the caravan respectively. The combination has an acceleration of 1.2ms^-2 with the engine exerting a constant driving force. Find: a) the driving force, b) the tension in the tow-bar. 
(g=10ms^-2)

loinducoeur · Answer

I understand that you have to add the resolved forces to get the driving force, but why do you add the frictional force to the driving force?
Also, why do you have to subtract the mgsin(theta) force from the F=ma force?

anonymous · Answer

@loinducoeur

I suggest you to draw a free body diagram in order to figure out what Fnet, Fexerted (or driving force) and the forces opposing motion are.

If you put a car toy in an incline it will go downward. That's because there is a component of its weight (w) exerting a force downward and parallel to the incline. This is wsin(θ) = mgsin(θ).

You know the acceleration of the car and the caravan and also know their respective masses, and so you're done with the Fnet. Once finding all the forces opposing the upward motion you can solve for the driving force and the tension.