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By the end of this section, you will be able to:
  • Define the property of pressure
  • Define and convert among the units of pressure measurements
  • Describe the operation of common tools for measuring gas pressure
  • Calculate pressure from manometer data

The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects ( [link] ). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.

The left side of this figure includes a graphic of the earth with an inverted rectangular prism extending from a point on it. Near the top of the image, the label, “square inch column of air molecules” is connected to the prism with a line segment. This label is also connected with a line segment to a downward pointing arrow at the right side of the figure. Beneath the arrow is a red circle labeled, “atmospheric pressure.” A narrow rectangle with a dashed line border extends from the bottom of the arrow vertically through the circle. Directly beneath this rectangle at the lower edge of the circle is a hand with a thumb appearing to be resting on a tabletop. The thumb is connected with a line segment to the label, “14.7 lbs of pressure on 1 square inch.” The red circle is sitting on top of the thumb.
The atmosphere above us exerts a large pressure on objects at the surface of the earth, roughly equal to the weight of a bowling ball pressing on an area the size of a human thumbnail.

Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.

In general, pressure    is defined as the force exerted on a given area: P = F A . Note that pressure is directly proportional to force and inversely proportional to area. Thus, pressure can be increased either by increasing the amount of force or by decreasing the area over which it is applied; pressure can be decreased by decreasing the force or increasing the area.

Let’s apply this concept to determine which would be more likely to fall through thin ice in [link] —the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in 2 ), so the pressure exerted by each foot is about 14 lb/in 2 :

pressure per elephant foot = 14,000 lb elephant × 1 elephant 4 feet × 1 foot 250 in 2 = 14 lb/in 2

The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in 2 , so the pressure exerted by each blade is about 30 lb/in 2 :

pressure per skate blade = 120 lb skater × 1 skater 2 blades × 1 blade 2 in 2 = 30 lb/in 2

Even though the elephant is more than one hundred-times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall though thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted:

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
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for grade 12 or grade 11?
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a rate of change in velocity of an object whith respect to time
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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