Static Equilibrium Question

Question

Question Description:

Three cables are connected at a central point B. The first cable runs from point A to point B at an angle 50˚ above the horizontal. Another cable runs from point B to point C. This cable has a horizontal dimension of 12 and a vertical dimension of 5 (creating a pythagorean triple). The final cable runs straight down from point B and is connected to a box, D, that wieghs 500lbs.

Find the tension in each cable.

My Answer(s) so far:

Tension in cable T_BD: 500lbs

Tension in cable T_AB: -698.846lbs (which doesn't make sense)

Tension in cable T_BC: 1003.580lbs

My Work:

I found the tension in cable T_BD to be the weight of the box since the ∑F (in the y direction) at box D is zero:

∑F = 0, therefore F_y = W = 500lbs

To find the tension of cables T_AB and T_BC I found up the following system of equations

T_BC*(5/13) + T_ABsin(50˚) = 500

T_BC*(12/13) + T_ABcos(50˚) = 0

The answer is supposed to be 500lbs for T_BD, 483.62lbs for T_AB, and 310.86 for T_BC

My mistake is somewhere in the system of equations but I am not sure what part is wrong. Thanks!

Answer 1

Let's go through what you did on this problem.

Tension in cable T_BD: 500lbs ← Yes! The cable must lift the box against gravity. Your reasoning that tension in the cable must balance the weight of the box is correct here.

One adjustment on your equations: Cables AB and BC are pulling in opposite horizontal directions. This means that the x-component of T_BC is the negative of the x-component of T_AB. This gets you

T_BC*(5/13) + T_ABsin(50˚) = 500

T_BC*(12/13) - T_ABcos(50˚) = 0

It looks like that is your only issue, and you should be able to solve this problem from there.