An electric pole shown in the figure below supports a power line that passes through it. A cable tied to the pole at B passes th
rough a hole in the struct at C and is then tied to the ground at B. There is no friction between the struct and cable. What is the force exerted by the cable onto the pole at B, if the tension in the cable is 47lb.
1 answer:
We have that The force exerted by the cable at point B on the pole is
From the Diagram
We can see that A,B.C is the area showing the circumference of a cable(power line)
Generally the equation for the Tension is mathematically given as
In conclusion
The force exerted by the cable at point B on the pole is
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Answer:
(a) f = 0.58Hz
(b) vmax = 0.364m/s
(c) amax = 1.32m/s^2
(d) E = 0.1J
(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)
Explanation:
(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:
(1)
k: spring constant = 20.0N/m
m: mass = 1.5kg
you replace the values of m and k for getting f:
The frequency of the oscillation is 0.58Hz
(b) The maximum speed is given by the following relation:
(2)
A: amplitude of the oscillations = 10.0cm = 0.10m
The maximum speed of the mass is 0.364 m/s.
The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.
(c) The maximum acceleration is given by the following formula:
The maximum acceleration is 1.32 m/s^2
The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.
(d) The total energy of the system is calculated with the maximum potential elastic energy:
The total energy is 0.1J
(e) The displacement as a function of time is: