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Ket [755]
3 years ago
15

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

n donde debe colocarse el punto de apoyo, hallar la VM​

Physics
1 answer:
Kitty [74]3 years ago
3 0

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

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Answer:

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A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?
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2 years ago
A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w
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Wow !  This one could have some twists and turns in it.
Fasten your seat belt.  It's going to be a boompy ride.

-- The buoyant force is precisely the missing <em>30N</em> .

--  In order to calculate the density of the frewium sample, we need to know
its mass and its volume.  Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

   Weight = (mass) x (gravity)
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   Mass = (185N) / (9.81 m/s²) = <u>18.858 kilograms of frewium</u> 

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³.  So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

           30N = (mass of the displaced water) x (9.81 m/s²)

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           Volume of displaced water = <u>3,058 cm³</u>

Finally, density of the frewium sample = (mass)/(volume)

      Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)

================================================

I'm thinking that this must  be the hard way to do it,
because I noticed that

       (weight in air) / (buoyant force) =  185N / 30N = <u>6.1666...</u>

So apparently . . .

        (density of a sample) / (density of water) =

                                  (weight of the sample in air) / (buoyant force in water) .

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3 0
3 years ago
Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4
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For E = 200 gpa and i = 65. 0(106) mm4,  the slope of end a of the cantilevered beam  is mathematically given as

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<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the   is mathematically given as

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In conclusion,  the slope is

A=0.0048rads

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