A 6.00-kg ornament is held at rest by two light wires that form 30° angles with the vertical,as shown in the figure.An external force of magnitude F acts vertically downward on the ornament.The tension exerted by each of the two wires is denoted by T.A free-body diagram,showing the four forces that act on the box,is shown in the figure.If the magnitude of force F is 410 N,what is the magnitude of the tension T?
A)271 N
B)235 N
C)188 N
D)376 N
E)470 N

Question

A 6.00-kg ornament is held at rest by two light wires that form 30° angles with the vertical,as shown in the figure.An external force of magnitude F acts vertically downward on the ornament.The tension exerted by each of the two wires is denoted by T.A free-body diagram,showing the four forces that act on the box,is shown in the figure.If the magnitude of force F is 410 N,what is the magnitude of the tension T? 
A)271 N 
B)235 N 
C)188 N 
D)376 N 
E)470 N

Quizplus · Accepted Answer

First, we need to resolve the external force F into its components. Since the wires are at an angle of 30° with the vertical, we can write:

F_vert = F cos 30° = (410 N)(√3 / 2) ≈ 355.9 N
F_horiz = F sin 30° = (410 N)(1 / 2) = 205 N

Now, we can write down the equations of equilibrium:

ΣF_vert = T1 + T2 - F_vert = 0
ΣF_horiz = - F_horiz = 0 (since there is no horizontal acceleration)

Since the ornament is at rest, there is no net force acting on it, so we can solve the above equations for T1 and T2:

T1 + T2 = F_vert
T1 = T2

Substituting the values, we get:

2T1 = (410 N)(√3 / 2)
T1 = T2 ≈ 135.5 N

Therefore, the magnitude of tension in each wire is approximately 135.5 N. The closest choice is A) 271 N, which is twice the tension in one wire.

A 600-Kg Ornament Is Held at Rest by Two Light Wires