Net force holds object in equilibrium

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Homework Help Overview

The problem involves a 6.0 kg box held at rest by two ropes forming 30-degree angles with the vertical, with an external force acting downward. The goal is to determine the magnitude of the tension in the ropes, given a downward force of 410 N.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the weight of the box and the net forces acting on it, questioning the correct approach to determine the tension in the ropes. There is a focus on how the two ropes contribute to balancing the downward forces.

Discussion Status

Some participants have provided insights into the balance of forces and the role of the two tensions in the vertical direction. There is ongoing exploration of the implications of the angles and the total forces involved, with no consensus reached yet.

Contextual Notes

Participants note confusion regarding the singular reference to force T despite the presence of two ropes, and there is a mention of the angle between the tensions being 60 degrees, which may affect the calculations. The original poster expresses a desire to understand the problem better without needing to submit a final answer.

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.)
 
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469 N force is balance by the components of 2 Ts in the vertical direction.
So 2*T*cos(30) = 469.
 
I think I understand what you mean by the balance of components, as the y-axis divides the two T forces, but . . . I'm still not following, exactly. Any other way to frame it? Would you have approached things similarly?
 
The title of the post says that Net force holds object in equilibrium
Since point of action of two tensions and the weight of the box and the force are in equilibrium, the resultant of two tensions must be equal to the total downward force.
The angle between tensions is 60 degrees. Find the resultant of the tensions and equate it to the total force. That gives you the magnitude of the tension.
 

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