Question

Determine the tension developed in cord DE required for equilibrium of the lamp. Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

Fde = 1080 N (3 sig fig)

Explanation:

Taking point E

Sum of forces at E in x direction:

Fde cos (30) - Fcd = 0  ..... Eq 1

Sum of forces at E in y direction:

Fde sin (30) - Wf = 0   ..... Eq 2

Wf = m*g = 55 * 9.81 = 539.55 N

Using Eq 2 to evaluate Fde

Fde = Wf / sin (30) = 539.55 / sin (30) = 1079.1 N

Answer: Fde = 1080 N (3 sig fig)