Bernouli's Equation/Pressure Problem

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Homework Help Overview

The problem involves a truck with a tarpaulin that bulges outward when the truck is in motion, specifically at a speed of 27 m/s. The context is centered around Bernoulli's equation and the pressure difference between the inside and outside of the tarpaulin, with air density given as 1.29 kg/m³.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Bernoulli's equation, questioning how to handle the unknown pressures and the velocity of the air inside versus outside the tarpaulin. There is consideration of the air inside the tarpaulin as stationary and the implications of the truck's motion on pressure differences.

Discussion Status

Some participants have offered insights into the relationship between the pressures and velocities, with one suggesting that the pressure difference can be calculated using Bernoulli's equation. There is acknowledgment of the assumption that the air inside the tarpaulin is at rest, which has led to further exploration of the pressure calculations.

Contextual Notes

Participants note that the problem involves horizontal motion, leading to the cancellation of certain terms in Bernoulli's equation. The discussion reflects uncertainty regarding the correct application of the equation and the interpretation of pressure conditions.

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Homework Statement


Consider a truck with a tarpaulin on top of it. When the truck is stationary the tarpaulin lies flat, but it bulges outward when the truck is speeding down the highway. The truck is traveling at 27 m/s. The density of air is 1.29 kg/m^3. by how much does the pressure inside the cargo area beneath the tarpaulin exceed the outside pressure

Homework Equations


bernoulis:
P(1) + 1/2pv^2 = P(2) + 1/2pv^2


The Attempt at a Solution


i plugged in the density and but with two unknown pressures and one unknown velocity i don't know how you would find the difference
 
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and i think the pgy's on either side of the equation cancel out because this is horizontal, so y=0.
 
I would consider the air inside the truck (under the tarpaulin) as stationary.
You can imagine the truck at rest and someone blowing some air on the outside.
 
When the truck is traveling at 27 m/s, the air above the tarpaulin is moving with the same velocity but in the opposite direction. The pressure inside this blowing air decreases. In side the tarpaulin air is at rest.
 
so if the air inside the tarpaulin can be considered stationary, you can use P(2) = P(1) + pgh. And the pgh would = 0, so the pressure inside would equal 1 x 10^5 Pa?
 
AH oh wait is this how you solve it:

P(1) + 1/2pv^2 = P(2) + 1/2pv^2
P(1) - P(2) = 1/2p(v2^2 - v1^2)
P(1) - P(2) = 1/2(1.29)(27^2 - 0^2)
P(1) - P(2) = 470 Pacorrect??
 
Correct.
 
thanks profusely
 

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