Question

In lecture, Dr. Selby used atmospheric air pressure to crush a can, by sucking the air out of it. (a) Calculate how many Newtons of force are exerted on the can. The dimensions of the un-crushed can are 25 cm × 18 cm × 11 cm. You may assume that all of the air was sucked out of the can. (b) Compare this force to your own weight. (c) Why don’t cans collapse all the time

Answer 1

Answer:

a) F1 = 1999.8 N , F2 = 4545 N ,  F3 = 2778 N , c) the cans do not collapse because the pressure is applied on both sides

Explanation:

Let's use the pressure equation

      P = F / A

Suppose we have atmospheric pressure 1.01 10⁵ Pa

Let's calculate the area of ​​the can that is a parallelepiped

Length L = 25 cm

width a = 18 cm

high h = 11 cm

Side area A = h a

      A = 11 18

      A1 = 198 10⁻⁴ m²

Lid area

     A2 = L a

     A2 = 25 18

     A2 = 450 10⁻⁴ m²

Other side area

    A3 = L h

    A3 = 25 11

    A3 = 275 10⁻⁴ m²

Now let's calculate the force on these sides

Side 1

     F1 = P * A1

     F1 = 1.01 10⁵ 198 10⁻⁴

     F1 = 1999.8 N

Side 2

    F2 = P A2

    F2 = 1.01 10⁵ 450 10⁻⁴

    F2 = 4545 N

Side 3

   F3 P A3

   F3 = 1.01 10⁵ 275 10⁻⁴

   F3 = 2778 N

We see that the force is greater on side 2 which is where the can should collapse

b) To compare the previous forces we must use the concept of density, in general the cans are made of aluminum that has a density of 2700 kg / m3

    d = m / V

    m = d * V

    V = L a h

    V = 0.25 0.18 0.11

    V = 0.00495 m3

    m = 2700 0.00495

    m = 13.4 kg

This is the maximum weight, because much of the volume we calculate is air that has a much lower density

   W = 13.4 * 9.8

   W = 131.3 N

Let's make the comparison by saying the two magnitudes

Side 1

    F1 / W = 1999.8 / 131.3

    F1 / W = 15.2

Side 2

    F2 / W = 4545 / 131.3

    F2 / W = 34.6

Side 3

   F3 / W = 2778 / 131.3

   F3 / W = 21.2

c) the cans do not collapse because the pressure is applied on both sides: outside and inside, so the net force is zero on each side.