Scott B. answered • 04/01/22
PhD in physics with one year experience as a professor
Since we're on an inclined surface, we'll want a tilted coordinate system. I'll call going up along the incline the positive parallel direction, and outward from the surface to be the positive perpendicular direction.
I'll call the weight of the wagon W, the angle of the incline theta, the coefficient of friction u, and the force of you pulling the wagon F.
We need to know the forces acting on the wagon, and their directions.
We've got the weight of the wagon, which annoyingly does not act in either the parallel or perpendicular directions. Whenever you have a vector that isn't pointing along one of the axes of your coordinate system, you want to replace it with the sum of two vectors that do. This is pretty common with these sorts of problems, so it's hopefully not too big a leap if I assert that we can replace the weight of the wagon with W cos(theta) acting into the surface (i.e. the negative perpendicular direction) and W sin(theta) acting down along the surface (the negative parallel direction).
Since the wagon is applying a force onto the surface of the inline, the incline must be pushing back with a force of its own. This is the normal force, which as the name implies, acts normal, or perpendicular, to the surface. Since this is up out of the surface, it's the positive perpendicular direction. I'll label this as N.
We've also got the applied force we're using to pull the wagon up the hill, F. F is acting to pull the wagon up along the surface of the hill, so it is in the positive parallel direction.
And finally we've got friction. Here you have to be really careful with the direction! In a lot of these inclined plane problems, the object (be it a wagon, a ball, or a block) is sliding down the incline, and so friction acts up along the surface, opposing the motion. But here, since we're pulling the wagon up along the surface, friction must act DOWN along the surface, again opposing the direction of motion. We know that the size of the friction force is uN.
I highly encourage you to sketch this all out to see it for yourself!
Now that we've got our forces, let's look at Newton's laws.
In the perpendicular direction, we've got it easy. The net force is N-W cos(theta), but since the block is not accelerating in the perpendicular direction, that net force must be 0. Hence, N=W cos(theta), a familiar result.
In the parallel direction, we've got a bit of legwork to do. The net force is easy enough, it's just F-uN-W sin(theta). We found N in the perpendicular calculation, so we can substitute that to get F-uW cos(theta)-W sin(theta), or F-W[u cos(theta)+sin(theta)] to simplify slightly. Since we want the minimum force necessary to pull the wagon up the hill, we want to set this to be 0 as well; at 0 net force, the acceleration is 0, meaning you could pull the wagon up at constant speed. Therefore,
F-W[u cos(theta)+sin(theta)]=0 -> F=W[u cos(theta)+sin(theta)]
Substituting the given values for W and theta, you should find that
F=45[0.12 cos(30)+sin(30)]=27.18 lbs
