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SSSSS [86.1K]
2 years ago
10

An elevator is accelerating upward at a rate of 3.6 m/s2. a block of mass 24 kg hangs by a low-mass rope from the ceiling, and a

nother block of mass 90 kg hangs by a low-mass rope from the upper block. (a) what are the tensions in the upper and lower ropes?
Physics
1 answer:
kozerog [31]2 years ago
5 0
The answer:

<span>When the elevator accelerates upward at a rate of 3.6 m/s², the value of the acceleration becomes

</span>A=g+3.6=13.4 m/s²
and by using the newton's law, F=mass x A, we have 
T1= (24 + 90 )x 13.4= 1527.6 N, where T1 is the <span>Tension in upper rope
</span> and 
T2= ( 90 )x 13.4= 1206N, where T2 is the Tension in lower rope

When the elevator accelerates downward at a rate of 3.6 m/s², the value of the acceleration becomes
A=9.8 - 3.6 = 6.2 m/s²

T1= (24 + 90 )x 6.2= 706.8 N, where T1 is the Tension in upper rope
 and 
T2= ( 90 )x 6.2= 558N, where T2 is the Tension in lower rope


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Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension in the string

\mu is the mass per unit length

For the string in the problem,

L = 30.0 m

\mu=9.00\cdot 10^{-3} kg/m

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz

The next frequencies (harmonics) are given by

f_n = nf

with n being an integer number and f being the fundamental frequency.

So we get:

f_2 = 2 (0.786 Hz)=1.572 Hz

f_3 = 3 (0.786 Hz)=2.358 Hz

f_4 = 4 (0.786 Hz)=3.144 Hz

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