- #1

Newlander

- 6

- 0

## 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/s

^{2}

v

_{i}= 0 m/s

f

_{F}= 410 N

f

_{T}= ?

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

F

_{net}= ma

cos [theta] = adjacent/hypotenuse

## The Attempt at a Solution

w = mg

f

_{W}= mg = 6.0 kg * -9.8 m/s

^{2}=-58.8 N

F

_{net}= ma

F

_{net, 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 f

_{w}and f

_{F}; 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.)