Hi guys! I'm new here. Here is the problem, verbatim: Two people pull to the right on a box with forces of 100 newtons at an angle of 60 degrees with respect to the x axis into the first quadrant and 80 newtons at an angle of 20 degrees down into the fourth quadrant respectively. A third person is to also pull on the box to exert a force that when the three forces are combined, the box moves to the right along the x axis. What is the magnitude and direction of the MINIMUM force the third person should exert to make this happen? My first source of confusion was this: I can find what the y-component of the third force is in order to zero any movement in the y-direction, but how do I know what x-component is required? I think I solved that problem by assuming that the problem assumes a frictionless surface, so the x-component of the force does not matter unless it is in the opposite direction. The forces on the box by persons 1. and 2. are in the x-direction anyways, so why would the third person need to exert a force in the x-direction? Is that reasonable or completely off? With that assumption made, I made this equation: V1y + V2y + V3y = 0 Substitute in the directions and magnitudes: 100N(sin60) + (-80N(sin30)) + (-V3y) = 0 Solve: V3y = 46.602 N and (I assume V3x = 0) When I go ahead and add x-components to find the vector of V1 + V2 + V3 = V4... 100(cos60) + 80(cos30) + 46.6(cos90) = 119.28 And then of course, V4 = √(119.28^2+0^2) = 119.28 So my V4 vector (the sum of the force on the box) is 119.28 N in the + x-direction. I finished the problem but I feel like it's wrong, but I can't think of any reason why it would be.