To solve this problem we will apply the concepts of equilibrium and Newton's second law.
According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So
![\sum F = ma](https://tex.z-dn.net/?f=%5Csum%20F%20%3D%20ma)
![T-mg = ma](https://tex.z-dn.net/?f=T-mg%20%3D%20ma)
Here,
T = Tension
m = Mass
g = Gravitational Acceleration
a = Acceleration (upward)
Rearranging to find T,
![T = m(g+a)](https://tex.z-dn.net/?f=T%20%3D%20m%28g%2Ba%29)
![T = (975)(9.8+0.754)](https://tex.z-dn.net/?f=T%20%3D%20%28975%29%289.8%2B0.754%29)
![T= 10290.15N](https://tex.z-dn.net/?f=T%3D%2010290.15N)
Therefore the tension force in the cable is 10290.15N
Answer:
Though the question is not specified here, but this information can determine the following quantity: period T= 6 secs, Frequency F=1/6 Hz, speed of rotation V= 2 pi ft/sec and wave length =pi/3 ft
Explanation:
Answer:
![r=0.41m](https://tex.z-dn.net/?f=r%3D0.41m)
Explanation:
Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.
![\tau=r\times F](https://tex.z-dn.net/?f=%5Ctau%3Dr%5Ctimes%20F)
Due to the definition of cross product, the magnitude of the torque is given by:
![\tau=rFsin\theta](https://tex.z-dn.net/?f=%5Ctau%3DrFsin%5Ctheta)
Where
is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when
is equal to one, solving for r:
![r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B%5Ctau%7D%7BF%7D%5C%5Cr%3D%5Cfrac%7B55%5Cfrac%7BN%7D%7Bm%7D%7D%7B135N%7D%5C%5Cr%3D0.41m)
We have that the spring constant is mathematically given as
![k=2.37*10^{11}N/m](https://tex.z-dn.net/?f=k%3D2.37%2A10%5E%7B11%7DN%2Fm)
Generally, the equation for angular velocity is mathematically given by
![\omega=\sqrt{k}{m}](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7Bk%7D%7Bm%7D)
Where
k=spring constant
And
![\omega =\frac{2\pi}{T}](https://tex.z-dn.net/?f=%5Comega%20%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D)
Therefore
![\frac{2\pi}{T}=\sqrt{k}{n}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Cpi%7D%7BT%7D%3D%5Csqrt%7Bk%7D%7Bn%7D)
Hence giving spring constant k
![k=m((\frac{2 \pi}{T})^2](https://tex.z-dn.net/?f=k%3Dm%28%28%5Cfrac%7B2%20%5Cpi%7D%7BT%7D%29%5E2)
Generally
Mass of earth ![m=5.97*10^{24}](https://tex.z-dn.net/?f=m%3D5.97%2A10%5E%7B24%7D)
Period for on complete resolution of Earth around the Sun
![T=365 days](https://tex.z-dn.net/?f=T%3D365%20days)
![T=365*24*3600](https://tex.z-dn.net/?f=T%3D365%2A24%2A3600)
Therefore
![k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2](https://tex.z-dn.net/?f=k%3D%285.97%2A10%5E%7B24%7D%29%28%28%5Cfrac%7B2%20%5Cpi%7D%7B365%2A24%2A3600%7D%29%5E2)
![k=2.37*10^{11}N/m](https://tex.z-dn.net/?f=k%3D2.37%2A10%5E%7B11%7DN%2Fm)
In conclusion
The effective spring constant of this simple harmonic motion is
![k=2.37*10^{11}N/m](https://tex.z-dn.net/?f=k%3D2.37%2A10%5E%7B11%7DN%2Fm)
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