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anyanavicka [17]

3 years ago

6

La superficie de unas botas suman 400 cm2 y la persona que las usa tiene 45 kg de masa, calcula la presión que ejerce sobre el p

iso. NOTA: Se debe encontrar la fuerza, para lo cual, se debe obtener el peso utilizando la fórmula de masa por gravedad que es 9.8 m / s2

1 answer:

liubo4ka [24]3 years ago

4 0

Answer:

11025 N / m²

Explanation:

Los siguientes datos se obtuvieron de la pregunta:

Área (A) = 400 cm²

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Presión (P) =?

A continuación, determinaremos la fuerza aplicada. Esto se puede obtener de la siguiente manera:

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Fuerza (F) =.?

F = m × g

F = 45 × 9,8

F = 441 N

A continuación, convertiremos 400 cm² a m². Esto se puede obtener de la siguiente manera:

1 cm² = 0,0001 m²

Por lo tanto,

400 cm² = 400 cm² × 0,0001 m² / 1 cm²

400 cm² = 0,04 m²

Por tanto, 400 cm² equivalen a 0,04 m².

Finalmente, determinaremos la presión ejercida de la siguiente manera:

Área (A) = 0.04 m².

Fuerza (F) = 441 N

Presión (P) =?

P = F / A

P = 441 / 0,04

P = 11025 N / m²

Por tanto, la presión ejercida es 11025 M / m²

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3 years ago

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Answer:

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The other forces acting on the yak are (see attached diagram):

* the force of gravity on the yak (identified in blue color in the image as $F_g$ ,

* the normal force (indicated in green in the image and identified by the letter "n") of the ledge on the yak as reaction to the yak's weight

* the force of static friction between the yak's hooves and the ledge (pictured in red in the image and identified with $f_s$ )

Since the normal force and the force of gravity on the yak cancel each other (balance - the yak is not moving vertically), the only forces we need to analyse are the force of the Tibetan's weight via the strap, and the force of static friction which should at least be equal in magnitude so the Tibetan doesn't fall. We assume these two forces are acting horizontally (one to the right: the Tibetan's weight, and one to the left: the static friction).

As we said, we want them to be at least equal so thy are in balance.

We recall that the force of static friction is the product of the normal force (n) times the coefficient of static friction ( $\mu_s$ ), such that: $f_s=\mu_s\,*\,n$

In our case these are the forces at play:

$F_g= M\,*\,g=450\, kg \,*\,g\\n=F_g=450\,kg\,*\,g\\f_s=\mu_s\,*\,n=\mu_s\,*450\,kg\,*\,g\\w=m\,*\,g=75\,kg\,*\,g$

So we need to find what is the minimum coefficient of static friction that precludes the Tibetan from falling. We therefore proceed to make an equality between the force of static friction on the yak and the weight of the Tibetan:

$f_s=w\\\mu_s\,*450\,kg\,*g=75\,kg\,*\,g$

and proceed to solve for the coefficient of friction by dividing both sides by "g" (which by the way cancels out), and by the yak's mass:

$\mu_s\,*450\,kg\,*g=75\,kg\,*\,g\\\mu_s=\frac{75}{450} \\\mu_s=0.1667$

where we have rounded to four decimal places the periodic number that the quotient generates. Notice that as expected, the coefficient of friction has no units (they all cancelled out in the division).

5 0

3 years ago