the system is in equilibrium ...
let = angle makes with the horizontal
= angle makes with the horizontal
you'll have to determine the values of the trig ratios from your given dimensions.
I'm having so much trouble with this and it's due in 3 days.
Here is what it looks like;
For W=175 pounds, computer tension force in the cables (both sides) for L=11 and X=(1,2,3,4,5). Which X position creates the greatest tension? Where is the best place to hang the weight?
I'm not asking for a solution, I just wonder if anyone can give me an equation I can use.
He didn't use the angles, he used the fact that the triangle formed by the horizontal and vertical forces has the same angles and is similar to the triangle formed by the roof, vertical, and cable. The horizontal force on each side, divided by the Tension in the rope on that side, is equal to the horizontal distance (x on the left, 10-x on the right) divided by the length of the hypotenuse ( on the left, on the right). Those must be equal in order that the weight must not move from side to side. That was the first of the two equations skeeter gave.
The vertical force on each side, divided by the tension in the rope on that side, is equal to the vertical distance, H, divided by the length of the hypotenuse ( on the left, on the right). The total of those two must be the weight itself. That is the second equation skeeter gave.