Pressure (SQA National 5 Physics): Revision Note
Exam code: X857 75
Pressure
Pressure is defined as:
Force exerted per unit area
The area should always be the cross-sectional area of the object
So that the force is at a right angle to the surface it is acting upon
The average pressure exerted on a surface can be calculated using the following relationship:
Where:
p = pressure measured in pascals (Pa)
1 pascal is equivalent to 1 newton per meter squared
1 Pa = 1 N m-2
F = force measured in newtons (N)
A = area measured in metres squared (m2)
The pressure relationship can be rearranged using a triangle
See Motion Equations for instructions on how to use a relationship triangle

Effects of changing variables
For a constant force:
Reducing the area increases the pressure
Increasing the area reduces the pressure
Pressure is inversely proportional to area
For a constant area:
Increasing the force increases the pressure
Decreasing the force decreases the pressure
Pressure is directly proportional to force

Examples of pressure
Pressure can be exerted by a solid on another solid
For example, the pressure exerted by a person standing on the ground

Pressure can be exerted by a gas on a surface
For example, when gas particles collide with the surfaces of a container

Air pressure
Air pressure arises due to the collisions of air particles with surfaces
The weight of the air above an object exerts a force at ground or sea level

Worked Example
The diagram below shows the parts of the lifting machine used to move the concert platform up and down.

The pump creates a pressure in the liquid of 5.28 × 105 Pa to move the platform upwards.
Calculate the force that the liquid applies to the piston.
Answer:
Step 1: List the known quantities
Cross-sectional area = 2.73 × 10-2 m2
Pressure = 5.28 × 105 Pa
Step 2: Write down the relevant relationship
Step 3: Rearrange for the force, F
Step 4: Substitute the values into the equation
F = (5.28 × 105) × (2.73 × 10-2) = 1.4414× 104
Step 5: Round to the appropriate number of significant figures and quote the correct unit
The least precise input value is 3 s.f.
Therefore, the answer can only be given to the same precision
F = 1.44× 104 N (3 s.f.)
Examiner Tips and Tricks
Exam questions will often use an unfamiliar context to make a question more difficult. These are called 'application questions'. It just means that you have to apply your understanding to a new situation. You already know the physics, you just have to think about how what you already know fits into the new situation.
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