Susie has a box that weighs 75N that is lying on top of a horizontal floor that is not frictionless. She pushes on the box with a force of 40N directed at an angle 41 degrees below the horizontal. Calculate the normal force that the floor exerts on the box. -------- We have not learned anything about friction yet, and it's not in the chapter, so I have no idea if there being friction matters or not for this problem. But ignoring that, it seems easy simple enough. I break the force exerted into the x and y components. 40*cos41 = 30.1 N 40*sin41 = 26.2 N x is pointed in the positive direction (+30.1N), and y is pointed in the negative direction (-26.2N). Since the box weighs 75N, it has a force of 75N downward, and that means the normal force is currently 75N upwards. Now, my question is, if everything I did so far is correct, is the new normal force just the total downward force (75N + 26.2N) = 101.2N, or is it equal to the sum of all the forces? Sqrt((75N+26.2N)^2+(30.1N)^2) = 105N?