Question

A trash can lid of mass 0.8 kg and area 0.25 m2
lies on top of a trash can but is not sealed shut in any
way. What minimum speed in m/s must air (density = 1.2 kg/m3
) blow over the top of the lid in order to
pop it off?

Answer 1

I at first thought that the force of gravity on it should be equal to the upward force but that leaves me with a velocity that isnt even one of the options.

Answer 2

Hi! I'm not sure if I'm correct, and I'm sorry if I cannot help. I saw this yesterday, too! I understand that the moving air has low pressure. So the force of pressure on the top of the lid will be less than force of pressure on the bottom. And the force diagram on the lid would look like

Answer 3

\(P=\dfrac{F}{A}\implies F=P\ A\), if that is what is relevant here. But that's all I know, and I don't see how we can find the pressures that we need too.

If you show your work I can help you check for math errors, but I can't check for conceptual errors at this time.

Maybe the \(P\)s can cancel. Like the pressure below is \(P\) and the pressure above is less than that by a factor of velocity, or just \(P\) is inversely proportional to \(v\).

Answer 4

Are there any equations you were given with \(P\) and \(\rho\) (density) that can help?

Answer 5

And \(v\), I meant!

Answer 6

I was right the answer key was wrong but thank you for the effort!

Answer 7

Well, that's good! I'm glad you got it right! :)

Answer 8

If you don't mind, could you explain what you had to do?

Answer 9

If it's too long, then don't bother! I was just wondering.

Answer 10

essentially the pressure difference should be equal to the weight of the lid per unit area, since the air is not moving in the trash can that side of bernouli's equation does not have a velocity, while the other side does so just take
P1+0=1/2(rho)v^2+P2
P1-P2=1/2(rho)v^2
P1-P2=(mg)/.25

Answer 11

divide by density and solve for v!

Answer 12

Cool! Thank you very much!