HW4 #3

I am getting questions about how to do #3 and the other tension problems. Here is a start; feel free to add more details if you think there is something missing here, or post a more specific question if this doesn't help quite enough.

A block is suspended from a long string. Since the block hangs from the middle and the angle on the right is the same as the angle on the left, it is reasonable to say that the tension T1 in the string up-and-to-the-left is the same as the tension T2 up-and-to-the-right (there is no reason why one of them would be a bigger force than the other).

The block is hanging from a vertical (up-down) string. Let's call the tension in the vertical string T. If you draw a force diagram for the block itself, you have Fg downward and T upward. From Newton's 2nd law and using up as your positive direction, ∑F=T-Fg=ma=0 and it =0 because the block is not accelerating. So T-Fg=0, and T=Fg=m*g, and you have a numerical answer for the tension T in the vertical string.

Now draw a force diagram (free body diagram) for the knot where all 3 strings attach. You have T downward, T1 up-and-to-the-left, and T2 up-and-to-the-right. Make a chart with columns for forces in the x direction, the y direction, and the "?" (idk) direction. T1 and T2 are in the ? direction, so figure out the x- and y-components of them. In question 2 you figured out the angle from the ceiling, so the horizontal (x) components of T1 is T1cosθ and T2 is T2cosθ, and the vertical (y) components are T1sinθ and T2sinθ.

Now you can use Newton's 2nd law in the vertical direction. If you choose up to be the positive direction again then ∑F=T1sinθ+T2sinθ-Fg=ma=0, and again it is =0 because the knot isn't accelerating. Since T1 is equal to T2 you can call them both T1, and solve for T1. That is the tension in the string.

Hope that helps. Good luck.