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aalyn [17]
3 years ago
6

A uniform metal meter-stick is balanced with a 1.0 kg rock attached to the left end of the stick. If the support is located 0.25

m from the left end, what is the mass of the meter-stick? (Hint: meter-stick is 1 m long). a) 0.6 kg b) 2.0 kg c) 1.0 kg d) 3.0 kg
Physics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

c) 1.0 kg

Explanation:

The mass of the stick will be located at the centre of the metre rule. Since the rock is located 0.25m from the pivot, the mass of the meter rule is also 0.25m to the Right of the support

According to law of moment

Sum of clockwise moment = sum of anti clockwise moments

Clockwise moment = M×0.25(mass of metre rule is M)

CW moment = 0.25M

Anti clockwise moment = 0.25×1

ACW moments = 0.25kgm

Equate;

0.25M = 0.25

M = 0.25/0.25

M = 1.0kg

Hence the mass of the metre rule is 1.0kg

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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
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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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