Question

An advertisement for a certain portable vacuum cleaner shows off its power with a photograph of the vacuum-cleaner wand suspending a bowling ball by the strength of its suction. The vacuum cleaner can maintain a moderate vacuum inside the apparatus at an absolute pressure of 3.53×104 Pa (against an outside atmospheric pressure of 1.01×105 Pa) when the intake wand is closed. The wand is a hollow metal cylinder with an inside diameter of 3.19 cm. What is the weight of the heaviest ball the vacuum cleaner can lift?

Answer 1

W = 52.5 N

Step-by-step explanation:

For this exercise let's use Newton's law and equilibrium

F₁ - F₂ - W = 0

Where the force F₁ is the external force applied by the air, the force F₂ is the internal force by the pressure in the vacuum and W is the weight of the ball

The pressure is given by

P = F / A

F = P A

A = π r²

r = d / 2

F = P π r²

Let's apply these equations to our case

F₁ = P_atm π r²

F₂ = P_ vacuum π r²

We substitute in the equilibrium equation

P_atm π r² - P_vacuum π r² - W = 0

W = (P_atm - P_vacuum) π r²

Let's calculate

r = 3.19 / 2 = 1,595 cm

r = 1,595 10⁻² m

W = (1.01 10⁵ -3.53 10⁴) pi (1,595 10⁻²)²

W = 0.657 10⁵ 7.992 10⁻⁴

W = 52.5 N