The following intervals would you expect T(tension) to fall : 2000 to 4000 N
The force acting on a system with static equilibrium is 0
[tex]\large{\boxed{\bold{\sum F=0}}[/tex]
(forces acting as translational motion only, not including rotational forces)
[tex]\displaystyle \sum F_x = 0\\\\ \sum F_y = 0[/tex]
We complete the questions:
A 20-kg loudspeaker is suspended 2.0 m below the ceiling by two ropes that are each 30? from vertical.
Find the value of T, the magnitude of the tension in either of the ropes.
Express your answer in newtons.
Suppose that the loudspeaker in the problem has a mass of 500 kg and the ropes hung 20? from the vertical. Into which of the following intervals would you expect T to fall? You don't have to calculate Texactly to answer this question; just make an estimate.
500 to 1000 N
1000 to 2000 N
2000 to 4000 N
4000 to 8000 N
8000 to 16,000 N
In a 500-kg loudspeaker system and two ropes that are each 20° from vertical, we see the forces acting on the y axis (vertical)
[tex]\displaystyle \sum F_y = 0\\\\T1_y+T2_y-w=0\\\\T1~cos~20^o+T2~cos~20^o=m\times g[/tex]
we assume g = 10 m/s², then :
[tex]\displaystyle 2T~cos~20=500\times 10\\\\T~cos~20=2500\\\\\boxed{\bold{T=26592 N}}}[/tex]
So the value of T is between 2000 and 4000 N
Newton's Law
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Keywords : the loudspeaker, ropes, Tension, mass, intervals
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We have that for the Question, it can be said that the range of the tension will be
T=2500N-4000N
From the question we are told
Suppose that the loudspeaker in the problem had a mass of 500 kg and the ropes hung 20∘ from the vertical. Into which of the following intervals would you expect t to fall? You do not have to calculate t exactly to answer this question; just make an estimate.
Generally the equation for the Tension is mathematically given as
[tex]T=\frac{mg}{2cos(\theta)}\\\\Therefore\\\\T=\frac{20*9.8}{2cos30}\\\\T=\frac{196}{1.73}N\\\\T=113N\\\\[/tex]
Therefore
The New Tension is mathematically given as
T'=25*113N
T'=2825N
Hence
The range of the tension will be
T=2500N-4000N
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