- #1
Newlander
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Homework Statement
"A 6.0 kg box is held at rest by two ropes that form 30-degree angles with the vertical. An external force F acts vertically downward on the box. The force exerted by each of the two ropes is denoted by T. . . . The magnitude of force F is 410 N. The magnitude of force T is closest to: . . ."
m = 6.0 kg
a = -9.8 m/s2
vi = 0 m/s
fF = 410 N
fT = ?
A figure is included in the practice example, but I am unable to upload it. Each rope is denoted by T--and this is key to the questions I have about the problem, noted below.
Homework Equations
w = mg
Fnet = ma
cos [theta] = adjacent/hypotenuse
The Attempt at a Solution
w = mg
fW = mg = 6.0 kg * -9.8 m/s2 =-58.8 N
Fnet = ma
Fnet, negative-y-axis = -58.8 N + (-410 N) = -469 N
cos [theta] = adjacent/hypotenuse = 469/T
--> T = 469/cos30 = 542 N
Concerns:
First, I'm not sure if I determined the magnitude of T the correct way. I figured I could use the positive y-axis as the equal but opposite force of the net force of fw and fF; this would in turn serve as the "adjacent" aspect of the trig function calculation.
Second, the problem refers to "force T" as if it is a singular force--but in the diagram (and explanation), there are two ropes denoted by T.
Third, the answer is provided: it is 271 N--exactly half of what I thought the magnitude of a *single* T is. I'm guessing I may be on the right path but am missing something because of the double nature of the force T--just not sure how to convey it in my work. (This isn't something to turn in--I just want to understand how to correctly do such a problem.)