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VladimirAG [237]
2 years ago
15

2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after

only two days, it had to be closed to traffic for safety reasons. On the opening day, in fact, so many people were crossing it at the same time that unexpected sideways oscillations of the bridge were observed. Further investigations indicated that the oscillation was caused by lateral forces produced by the synchronization of steps taken by the pedestrians. Although the origin of this cadence synchronization was new to the engineers, its effect on the structure of the bridge was very well known. The combined forces exerted by the pedestrians as they were walking in synchronization had a frequency very close to the natural frequency of the bridge, and so resonance occurred.
Consider an oscillating system of mass m and natural angular frequency ωn. When the system is subjected to a periodic external (driving) force, whose maximum value is Fmax and angular frequency is ωd, the amplitude of the driven oscillations is A=Fmax(k−mωd2)2+(bωd)2√, where k is the force constant of the system and b is the damping constant. We will use this simple model to study the oscillations of the Millennium Bridge.
Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 N at a frequency of about 1 Hz. Given that the mass of the bridge is about 2000 kg per linear meter, how many people were walking along the 144- m-long central span of the bridge at one time, when an oscillation amplitude of 75 mm was observed in that section of the bridge? Take the damping constant to be such that the amplitude of the undriven oscillations would decay to 1/e of its original value in a time t=6T, where T is the period of the undriven, undamped system.
Physics
1 answer:
sweet [91]2 years ago
6 0

Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

A(t)=A_oe^{\frac{-bt}{2m}}

or amplitude for unforce oscillation

\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}

or

\frac{6bt}{2m}=1

or

b=\frac{m}{3T}

Now, provided in the question Amplitude of the driven oscillation

A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}

the value of the maximum amplitude is obtained (k=m\omega_d^2)

thus,

A=\frac{F_{max}}{(b\omega_d}

Now, for n people on the bridge

Fmax = nF

thus,

max amplitude

0.075=\frac{nF}{((\frac{m}{3T})2\pi}

or

n = 1810

hence, there were 1810 people on the bridge

b)A=\frac{F_{max}}{(b\omega_d}

since the effect of damping in the millenium bridge is 3 times

thus,

b=3b

therefore,

A=\frac{F_{max}}{(3b\omega_d}

or

A=\frac{1}{(3}A_o

or

A=\frac{1}{(3}0.075

or

A = 0.025 m = 25 mm

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A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
dybincka [34]

Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
3 years ago
un columpio de balancin tiene una barra de 6m de longitud y en ella se sientan 2 personas,una de 60kg y otra de 40kg, calcular e
Kitty [74]

Answer:

<em>El punto de apoyo debe colocarse a 3.6 metros de la persona de 40 Kg</em>

<em>La ventaja mecánica es 1.5</em>

Explanation:

<u>Máquinas Simples</u>

Un balancín es un ejemplo de máquina simple, donde se aplica una fuerza física y ésta puede amplificarse o reducirse a voluntad cambiando la configuración física de la máquina.

En nuestro caso, el punto de apoyo o fulcro se coloca entre las dos fuerzas constituyendo una máquina de primer grado.

La situación planteada se muesta en la figura anexa. Debemos averiguar el valor de x para que las dos personas sentadas en el balancín puedan estar en equilibrio.

Para determinar el valor de x, se establece la condición de equilibrio de torques mecánicos. Ya que el balancín se asume en reposo, los torques aplicados de cada lado del mismo deben ser iguales, haciendo que el torque neto sea cero.

El torque es el producto de la fuerza por la distancia:

T = F.d

De cada extremo del balancín, se aplica una fuerza igual al peso de cada persona, es decir, llamando F1 al peso de la persona de 40 Kg y F2 al peso de la persona de 60 Kg:

F_1 = 40 kg * 9.8 m/s^2=392N\\\\F_2 = 60 kg * 9.8 m/s^2=588 N

El torque neto del balancín debe ser cero, es decir (refiérase a la figura):

F_1*x=F_2*(6-x)

Reemplazando los valores obtenidos:

392*x=588*(6-x)

Operando:

392*x+588*x=588*6

Simplificando

980*x=3528

Resolviendo

\displaystyle x=\frac{3528}{980}=3.6\ m

El punto de apoyo debe colocarse a 3.6 metros de la persona de 40 Kg, es decir, a (6 - 3.6) = 2.4 metros de la persona de 60 Kg

La ventaja mecánica se calcula como el cociente de ambas distancias

\displaystyle VM=\frac{3.6}{2.4}=1.5

La ventaja mecánica es 1.5, es decir, se amplifica la fuerza vez y media

3 0
3 years ago
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