Example: A rocket thrust is applied and the rocket moves forward at the rate of thrust. The rocket will increase its speed as long each second as long as the thrust is applied (assuming there is no counter force of friction)
Pressure, Force and Area
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There is a relation between pressure, force and area. To understand this relationship we must first understand each variable. Force is the measure in Newton’s “N”.
1. Force is describe as a Newton which is define as the measure as the force required to move a one kilogram object at an accelerated rate one meter per second for each second that the force is being exerted.
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Example: A rocket thrust is applied and the rocket moves forward at the rate of thrust. The rocket will increase its speed as long each second as long as the thrust is applied (assuming there is no counter force of friction)
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If the rocket weighed 100 kg and the force of gravity is 9.8 m/per second2 the total force the rocket would exert prior to take off would be 980 N. The rocket thrust would have to be greater than 980 N in order for the rocket to lift off the pad!
2. Area is described as length X width:
Example if you have a field that measures 50 cm by 100 cm (I know it’s a really small field!) then to find the total area of the field inside you would use the following equation:
A= L x W
A = 50 cm X 100 cm
A= 5000 cm2
Pressure is described as a relationship of force over area.
This can be shown mathematical relationship: P= F/A
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Example If you have an 100 kg rock exerts a 980 N over an area of 100 cm2 what is the pressure?
P= 980N/100cm2 P=9.8 N/cm2
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Let’s say you reduce the area in which the rock sits to 50 cm2 what will happen to the pressure?
P=980 N/50 cm2 P=19.6 N/cm2
If you said the pressure would increase you are correct! |
In both cases the mass and force remains constant while there is an increase in pressure because you reduce the area in which the force is spread out!
Remember that you can change the mathematical relationship to find any of the three variables, as long as you know two of the variables!
P= F/A F = PA A= F/P
Let’s do some problems!
What is the force in Newton’s of an object that exerts a pressure of 1 atm over an area 10 cm2?
P= 1 atm
A= 10 cm2
F= PA
F= 1 atm X 10 cm2
F= 10 N
Find the Area in cm2 of a surface whose force is 15 N and pressure is 2 atm.
F= 15 N
P= 2 atm
A= F/P
A= 15 N/ 2 atm
A= 7.5 cm2
What is the pressure in N/cm2 of an object that exerts 25 N over 30 cm2?
F= 25 N
A= 30m2
P=F/A
P= 25 N/30 cm2
P= .83 N/cm2
