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Outline - Lectures weeks 9-12
Chapter 6: Balance in nature- description of energy balance in the atmosphere
Chapter 7: Properties of water- surface tension, pressure, specific heat capacity
Chapter 8: Materials: Elasticity & Viscosity- stress and strain, rheology
Chapter 9: Farm machinery: Friction & Lubrication- friction
Chapter 10: Farm machinery: Stability- Newton’s laws, torque
Chapter 11: Farm machinery: Vibrations- oscillations, resonance
Chapter 10:Farm machinery: Stability
Direction of friction revisited
From University PhysicsYoung & Freedman
Newton’s laws of motionFirst law
• Push and release a puck on a table: it slides and then comes to a halt
- reduce the friction (waxed surface, or air table): the puck slides further
- in the limit of zero friction, the puck would slide indefinitely, at constant speed
- once the puck is released, there is no net force acting on it (in the frictionless case)
First law: A body acted on by no net forces moves with constant velocity (which may be zero)
• Forces are vector quantities: you need to add them vectorially
Lift force L
Weight force W
Thrust T Drag D
• If the aircraft is cruising at a constant velocity, the four forces sum vectorially to zero
- we write L + T + D + W = 0 or `ΣFi = 0’
From http://www.drnicolemunk.de/Bilder/Airbus-A380-Emirates-Airlines-Flug-Dubai-Burj-Al-Arab-Munk.jpg
Second law
Second law: A body subject to a net force accelerates(changes its velocity) according to ΣFi = ma where m isthe mass, a is the acceleration
From University Physics, Young & Freedman
• If ΣFi = 0 then a = 0, i.e. v = const (first law)
Third law
• Forces involve the interaction of two bodies, e.g. your hand and the puck
- you push on the puck, and it pushes back on you- experiment shows these forces are equal and opposite
Third law: For every force applied to an object, the object exerts an equal and opposite force on the body applying the force
• The forces are called action-reaction pairs• The forces act on different objects
- hence they don’t `cancel out’
• The puck pushes down on the table with the force W• The table pushes back on the puck with the normal reaction force N• The forces are equal and opposite
- they act on different objects
Donkey and cart problem
Question: The donkey pulls on the cart. By Newton’s third law, the cart pulls on the donkey with an equal and opposite force. Why then does the cart start to move?
Answer: The two forces act on different objects.To determine the motion of the cart, we consideronly the forces acting on the cart. If there is a netforce on the cart, it will accelerate (Second law).
Force on cart due to horseFrictional forces
• If the size of the force due to the horse exceeds the frictional forces, the cart accelerates
Torque
F
x • For a wheel rotating about an axle as shown, |τ| = Fx is the magnitude of the torque about the axle
• Torque is the rotational analog of force• The torque τ of a force F about a given axis is the vector cross product τ = r x F
- r is the position vector of the point of application of the force wrt the axis
• A non-zero net torque about an axis produces an angular acceleration, i.e. changes the angular speed of rotation (analog of second law)
W
F
F A
• Consider a wheel suspended by its axle and free to rotate. A mass is hung on a string wrapped around the wheel and released.• The forces on the wheel are shown• There is a net torque about the axle due to the force F• There is no net torque due to W or FA about the axle - why?• The wheel starts to rotate due to this net torque
Mass
Suspending force
Weight of wheel
Forcedue toweightofmass
Wheel
Axle
Equilibrium• General motion of a rigid body: translation + rotation
• Equilibrium: motion does not change with time• Newton’s second law: equilibrium requires that the net forces are zero, and the net torques are zero
Translational equilibrium (net force = 0): ΣFi = 0 Rotational equilibrium (net torque = 0): Στi = 0
Torque
Fx
W
F
F A
Forces Torques
F
F
F
W
A
Fx
-Fx
Forces Torques
• Example 1:
- Net forces zero- Net torque about axle non-zero- Wheel stays in position but starts to rotate
• Example 2:
- Net forces zero- Net torque about axle zero- Wheel is in equilibrium
Centre of mass (centre of gravity)
• Divide a plane rigid body into mass elements mi
- M = Σ mi is total mass• The torque about an axis O due to gravity g is
τ = Σ ri x mig = (M-1 Σ mi ri) x Mg = rcom x W
• The net torque is equivalent to that due to the weight W acting through the centre of mass rcom = M-1 Σ mi ri
- also called centre of gravity (for constant g) • To locate the centre of mass of a plane body
- suspend an object, draw a line down through point of suspension (rcom is on this line)- suspend by a different point, repeat- intersection of lines is rcom
Centre of mass and tipping over
From University Physics, Young & Freedman
• A body supported at several points must have its centre of mass within the area of support, as shown
- otherwise there is a net torque about one of the points of support: it will tip
Centre of mass and tipping over
• Stacking: the centre of mass of all objects above must be within the base of support, for each object
- possible for top object to be outside the base of the bottom!
Agricultural applications
• When towing objects, attach rope at a low point, so that the torque about the base of support does not tip the object
• To increase stability of loads (e.g. on trailers)- widen the base of support- lower the centre of mass
F
W
P
Summary
• Newton’s three laws of motion • Torque• Equilibrium• Centre of mass• Stability
