This is not HW problem specific, but more of a lack of understanding for an equation in answer.
A pair of students are lifting a heavy trunk on move-in day. (Figure 1) Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity v⃗ . Each rope makes an angle θ with respect to the vertical. The gravitational force acting on the trunk has magnitude FG.
(the figure is just 2 people pulling rope on either side of the trunk)
What is the force of tension in each rope? The answer is T = (FG) / (2cos(θ))
The Attempt at a Solution
I undertand that the y-components of the tensions will be T_y = T*cosθ (because θ is here measured from the vertical)
I also get that F_mg is the downward force which points along negative-y axis.
I also get that all forces must balance because it says constant v, so no acceleration. So mg + T_y = 0 and T_1x + T_x2 = 0
But Im still not understanding how the tension in each rope, T = (FG) / (2cos(θ))? Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?
thanks for any help