Principles & Processes in Agriculture- once the puck is released, there is no net force acting on it...
Transcript of Principles & Processes in Agriculture- once the puck is released, there is no net force acting on it...
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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
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Chapter 10:Farm machinery: Stability
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Direction of friction revisited
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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
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- 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
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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
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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)
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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’
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• 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
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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?
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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
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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)
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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
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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
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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
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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
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• 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
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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
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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!
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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
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Summary
• Newton’s three laws of motion • Torque• Equilibrium• Centre of mass• Stability