Net force holds object in equilibrium

AI Thread Summary
A 6.0 kg box is held in equilibrium by two ropes at 30-degree angles with the vertical while a downward force of 410 N acts on it. The weight of the box is calculated as 58.8 N, leading to a total downward force of 469 N when combined with the external force. The tension in each rope, denoted as T, must balance this total downward force. The correct approach involves recognizing that the vertical components of the two tensions must equal the total downward force, leading to the equation 2*T*cos(30) = 469 N. The final answer for the magnitude of a single tension T is 271 N, confirming the need to account for both ropes in the calculations.
Newlander
Messages
6
Reaction score
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/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.)
 
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top