AQA Science © Nelson Thornes Ltd P3 1.2 What are we studying today? AND + not mine but… pressyour PRESSURE high drawerlicks HYDRAULICS

AQA Science © Nelson Thornes Ltd P ?feature=endscreen&v=qiFL_ yOvV8Y&NR=1 Siri’s feeling the pressure

AQA Science © Nelson Thornes Ltd P3 1.2 Pressure and Hydraulics Objectives Be able to describe what pressure is and how it applied to hydraulic systems. 08 September 2015

P3 1.2 AQA Science © Nelson Thornes Ltd Outcomes All MUST be able to describe pressure is. MOST Should be able to use the equation for pressure. SOME could be able to rearrange the equation for pressure and use it in different situations.

P3 1.2 Definition: Hydraulics is the use of pressure in liquids. To understand hydraulics, first we must understand pressure. Pressure is the Force per Unit Area. Pressure (Pascals) = Force (N) / Area (m 2 ) Crushing Can demo:

P3 1.2 How a gas exerts pressure A gas consists of molecules in constant random motion. When a molecule collides with a surface it reverses direction due to the force exerted on it by the surface. The molecule in turn exerts a force back on the surface. The pressure exerted by the gas is equal to the total force exerted by the molecules on a particular area of the surface divided by the area. pressure = force / area

P3 1.2 Pressure examples pressure in Pa or N/m 2 Space (vacuum)0 Air pressure at the top of Mount Everest Average pressure of the Earth’s atmosphere at sea level at 0°C Typical tyre pressure Pressure 10m below the surface of the sea Estimated pressure at the depth (3.8km) of the wreck of the Titanic

P3 1.2 Other pressure units Note: You do not need to learn any of these for the GCSE exam Atmospheres (atm) Often used to measure the pressure of a gas. An atmosphere is the average pressure of the Earth’s atmosphere at sea-level at a temperature of 0°C. Standard atmospheric pressure = Pa (about 101 kPa) Bars and millibars (bar; mbar) Also used to measure gas pressure. One bar is about the same as one atmosphere. Millibars are often found on weather charts millibars = 1 bar = 100 kPa

P3 1.2 Pounds per square inch (psi) Often used to measure car tyre pressures. 1 psi = 6895 Pa 1 atm = 101 kPa = 14.7 psi Inches of mercury (inHg) Often found on domestic barometers. 1 inHg = 3386 Pa 1 atm = 101 kPa = 29.9 inHg Examples: Fair weather – high pressure: 30.5 inHg Rain – low pressure: 29.0 inHg tyre pressure gauge

P3 1.2 Pressure and Hydraulics AQA Science © Nelson Thornes Ltd Watch: BUxv0qE&playnext=1&list=PL1696A54 29EF5BC33&feature=results_videohttp:// BUxv0qE&playnext=1&list=PL1696A54 29EF5BC33&feature=results_video Prepare to describe the 3 most important things you learn from this – you may want to jot points down in the back of your book

P3 1.2 Pressure, p pressure = force area p = F A units: force, F – newtons (N) area, A – metres squared (m 2 ) pressure, p – pascals (Pa)

P3 1.2 also: force = pressure x area and: area = force pressure p A F Note: 1 Pa is the same as 1 newton per square metre (N/m 2 )

P3 1.2 Pressure exerted by a person on a floor 1. Weigh the person in newtons. This gives the downward force, F exerted on the floor. 2. Draw, on graph paper, the outline of the person’s feet or shoes. 3. Use the graph paper outlines to calculate the area of contact, A with the floor in metres squared. (Note: 1m 2 = cm 2 ) 4. Calculate the pressure in pascals using: p = F / A

P3 1.2 Typical results 1. Weight of person: _____ N 2. Outline area of both feet in cm 2 ____ 3. Outline area of both feet in m 2 _____ 4. Pressure = ________ = _______ Pa N m

P3 1.2 Why off-road vehicles have large tyres or tracks In both cases the area of contact with the ground is maximised. This causes the pressure to be minimised as: pressure = vehicle weight ÷ area Lower pressure means that the vehicle does not sink into the ground.

P3 1.2 Question sheet – 15 minutes Work with your partner and complete as many questions as you can Extension on next page

P3 1.2 Pressure exerted by a block question - extension The metal block, shown opposite, has a weight of N. Calculate the maximum and minimum pressures it can exert when placed on one of its surfaces. Maximum pressure occurs when the block is placed on its smallest area surface (2m x 3m) p = F / A = N / 6m 2 Maximum pressure = Pa Minimum pressure occurs when the block is placed on its largest area surface (3m x 5m) p = F / A = N / 15m 2 Minimum pressure = Pa 2m2m 5m5m 3m3m

P3 1.2 Question 1 Calculate the pressure exerted by a force of 200N when applied over an area of 4m 2. p = F / A = 200N / 4m 2 pressure = 50 Pa

P3 1.2 Question 2 Calculate the force exerted by a gas of pressure Pa on an object of surface area 3m 2. p = F / A becomes: F = p x A = Pa x 3 m 2 force = N

P3 1.2 Question 3 Calculate the area that will experience a force of 6000N from a liquid exerting a pressure of 300kPa. p = F / A becomes: A = F / p = 6000 N ÷ 300 kPa = 6000 N ÷ Pa area = 0.02 m 2

P3 1.2 Complete: forceareapressure 40 N8 m 2 Pa 500 N20 m 2 25 Pa 400 N5 m 2 80 Pa 20 N2 cm kPa 6 N2 mm 2 3 MPa

P3 1.2 What can you do? I most definitely can describe what pressure is. I should be able to apply my understanding and use the equation for pressure. I may be able to rearrange the equation for pressure and use it in different situations. Tell me what you thought was easy about this lesson and what was difficult. Anything that would help you in the future? Eg: I thought that …. was easy/difficult because … It would help if …