Applying Newtons laws ( Equilibrium)

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SUMMARY

The discussion focuses on calculating the tension in two cables supporting a 26.21 kg loudspeaker suspended 1.15 m below the ceiling. The cables, each 3.91 m long, form equal angles with the vertical, determined to be approximately 17 degrees. The tension in each cable is calculated using the formula T = (1/2 * weight) / sin(θ), where θ is the angle of the cables. The vertical component of the tension must balance the weight of the speaker, confirming that the horizontal components are equal and opposite, ensuring equilibrium.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic trigonometry, specifically sine and cosine functions
  • Ability to resolve vector components
  • Knowledge of weight calculation using mass and gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Learn about vector resolution in physics
  • Study the application of Newton's laws in static equilibrium
  • Explore trigonometric functions in real-world applications
  • Investigate tension calculations in various mechanical systems
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Students studying physics, particularly those focusing on mechanics and equilibrium, as well as educators seeking to enhance their teaching methods in these topics.

sktgurl930
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Homework Statement


A 26.21 kg loudspeaker is suspended 1.15 m below the ceiling by two 3.91-m-long cables that angle outward at equal angles. What is the tension in the cables?

Homework Equations


I have no idea


The Attempt at a Solution


well i know that i have the weight which is 26.21 kg and how far its is from the ceiling
and that I am trying to find N but i just don't kno wat equation to use but once i find N would i have to divide by 2 since there's 2 cables??
 
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sktgurl930 said:

Homework Statement


A 26.21 kg loudspeaker is suspended 1.15 m below the ceiling by two 3.91-m-long cables that angle outward at equal angles. What is the tension in the cables?

Homework Equations


I have no idea

The Attempt at a Solution


well i know that i have the weight which is 26.21 kg and how far its is from the ceiling
and that I am trying to find N but i just don't kno wat equation to use but once i find N would i have to divide by 2 since there's 2 cables??

Figure what angles the cables make holding the speaker first.

Then draw your vector diagram of the tensions and resolve the x and y components. You know the X direction components must add to 0 and the Y components muust add to the 26.21 kg.
 
ok so would the angles be 17 since they are equal angles??

and i got 17 by making a right triangle with 3.91 being the hypotenuse, and since the speaker is 1.15m away from the ceiling making that one of the sides of the triangle with the right angle.

im i right?? or totally off
 
im not going to lie I am teacher isn't very good at teaching
so i have no idea wat ur talkin about with x and y

please don't hate me
 
sktgurl930 said:
ok so would the angles be 17 since they are equal angles??

and i got 17 by making a right triangle with 3.91 being the hypotenuse, and since the speaker is 1.15m away from the ceiling making that one of the sides of the triangle with the right angle.

im i right?? or totally off

That would be correct. It is 17 degrees. I get 17.1, but that's close enough.

So now you know that 1/2 the weight 26.21kg * 9.8 m/s2 is carried by the vertical component of the tension in the cable. As you noted each cable will carry 1/2.

If that vertical force must be carried in the cable then the tension must be = 1/2 weight/Sine(17).
(Answer in Newtons)

That treats the Tension as having 2 force components. One vertically (you found) and the other horizontally (given by 26.21*(Cos(17)). When added together as you would add vectors, that adds to the answer you just found. Since the speaker is not in motion, you know that the horizontal components are equal and opposite.

Another way to find the answer would be to recognize that

9.8*26.21*(1/2)/1.15 = Total Tension/3.91

This would be because the definition of Sine is Vertical over hypotenuse and those forces must remain in the same ratio, so ... you can exploit that as the quick way to calculate it as like ratios.
 
thank u and i wish u were my teacher dude ur a life saver!
 

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