How Does Tire Pressure Relate to Automobile Weight?

[imnotsmart]

The four tires of an automobile are inflated to a gauge pressure of 2.2*10^5 Pa. Each tire has an area of 0.024 m2 in contact with the ground. Determine the weight of the automobile.
How do you go about finding the answer to this problem?

 

[xanthym]

The four tires of an automobile are inflated to a gauge pressure of 2.2*10^5 Pa. Each tire has an area of 0.024 m2 in contact with the ground. Determine the weight of the automobile.
How do you go about finding the answer to this problem?

SOLUTION HINTS:
{Weight of Auto} = {Number of Tires}*{Tire Gauge Pressure (ea tire in Pa = N/m^2)}*{Contact Area (each tire in m^2)}
Answer should be approx (21,120 N).


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[imnotsmart]

What about this problem..
A 81 kg man in a 4.0 kg chair tilts back so that all the weight is balanced on two legs of the chair. Assume that each leg makes contact with the floor over a circular area with a radius of 1.0 cm, and find the pressure exerted on the floor by each leg.

 

[xanthym]

What about this problem..
A 81 kg man in a 4.0 kg chair tilts back so that all the weight is balanced on two legs of the chair. Assume that each leg makes contact with the floor over a circular area with a radius of 1.0 cm, and find the pressure exerted on the floor by each leg.

SOLUTION HINTS:
Use equation below to calculate required pressure. Remember to convert chair leg circle Radius from "cm" to "m".

\mbox{(Pressure Ea Chair Leg in N/m^2)} \ \ =

= \ \ \frac {\displaystyle \left \mathbf{\{} \mbox{(Mass of Man in kg)} + \mbox{(Mass of Chair in kg)} \right \mathbf{\}} \cdot \mbox{(9.81 m/sec^2)} } { \pi \cdot \mbox{(Ea Chair Leg Contact Circle \underline{Radius} \color{red}\textbf{in m}\color{black})}^{2} \cdot \mbox{(Number Chair Legs in Contact)} }


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[imnotsmart]

so the radius squared is .0001 and the number chair legs in contact is two...correct?

 

[xanthym]

so the radius squared is .0001 and the number chair legs in contact is two...correct?

Correct.
(R^2 = 0.0001 m^2)
(# Chair Legs in Contact = 2)


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[imnotsmart]

Ok, I figured that one out. This is the last one that I need help with if you don't mind.
Water is to be pumped to the top of a building which is 850 ft high. What gauge pressure is needed in the water line at the base of the building to raise the water to this height?

 

[xanthym]

Ok, I figured that one out. This is the last one that I need help with if you don't mind.
Water is to be pumped to the top of a building which is 850 ft high. What gauge pressure is needed in the water line at the base of the building to raise the water to this height?

SOLUTION HINTS:
Use equation below to compute pressure (in N/m^2). Remember to convert Height from "ft" to "m".
{Gauge Pressure (in N/m^2)} = {Water Density (in kg/m^3)}*(g)*{Height (in m)} =
= (1000 kg/m^3)*(9.81 m/sec^2)*{Height (in m)}


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[imnotsmart]

thanks for all your help buddy

 

[imnotsmart]

I have two more questions for anyone can help...
1. A frog in a hemispherical pod finds that he just floats without sinking in a fluid of density 1.25 g/cm3. If the pod has a radius of 9.00 cm and negligible mass, what is the mass of the frog?
2. A cowboy at a dude ranch fills a horse trough that is 1.6 m long, 60 cm wide, and 40 cm deep. He uses a 1.9 cm diameter hose from which water emerges at 1.4 m/s. How long does it take him to fill the trough?

Just need some direction...no answer.

 

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how do you messure pressure that you exert on the floor when you stand on one foot ?