A farm tractor pulls a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer (ignore friction)?




My attempt:

I used F=ma to find that the force down (using gravity-9.8m/s^2 for a) so that the force directly down is 36260 N
Then to find the hypotenuse (parallel force) shouldn't I use equation sin 18 = 36260/F ?
It says the correct answer is 11kN.

So it should actually be Sin (18) = F/36260

Why would F be opposite and 36260 be the hypotenuse?
I thought the 36260 represents straight downward force of gravity and F is the parallel force to the surface

I believe if you drew a force diagram it would help you better, but i havent done physics in years. so 36260 is the hypotenuse since it will be the largest value. im looking for a pic atm that should help you understand



Heres another one



they just use some math properties to create a triangle with the relevant forces.

This post was edited by cialda on Oct 6 2013 04:29pm

The magnitude of the parallel component (F) of the weight (w = mg) is the opposite side of a right triangle (see #4);

F = m·g·sin('theta') = ...

F is directed down the incline (opposite the direction of motion).
Apply Newton's Second Law to the trailer (which is being pulled at constant speed), to get the result.

This post was edited by Quorra on Oct 7 2013 09:19am