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8 - Newton's Laws of Motion (8.2 Using F = ma (Lift problems (T-mg=ma,…
8 - Newton's Laws of Motion
8.1 Force and Acceleration
The push force is opposed by friction and as soon as the object stops moving it gets stopped by friction
Newtons first law of motion
Objects either stay at rest or moves with constant velocity unless acted on by a force
Object moving at constant velocity is either acted on by no forces or the forces acting on it are balanced(resultant force = 0)
Investigating force and motion
To show how the velocity of an object changes if it is acted on by a constant force a dynamics trolley and a motion sensor can be connected to a computer to show a graph on how the velocity of the trolley changes with time
The trolley is pulled along a sloping runway using one or more elastic bands stretched at the same length. The trolley should move at constant speed down the runway.
As a result a velocity time-graph should show an increase of constant rate.Acceleration therefore should be constant and can be worked out from the gradient.
The results should show that
F is proportional to ma
or
F=ma
which is
Newton's second law
Weight
W = mg
In equilibrium, support force is equal and opposite to weight
g is also referred as the
gravitational field strength
at a given position(Earth's is 9.8)
The mass of an object is a measure of it
inertia
, which is its resistance to change of motion
8.2 Using F = ma
Two forces in opposite directions
Resultant force, F1 - F2 = ma
a is the acceleration which is in the same direction as F1
Towing a trailer
Car is subjected to driving force F pushing it forwards and the tension T in the tow bare holding it back. Therefore resultant force = F-T=Ma
The force on the trailer is due to the tension T in the tow bar pulling it forwards. Therefore T=ma
Combining the two equations gives
F=Ma+ma =(M+m)a
Rocket problems -
Rocket thrust T=mg+ma
Lift problems
T-mg=ma
If lift is moving at constant velocity, then a=0 so T=mg
If lift is moving up and accelerating, then a>0 so T=mg+ma>mg
If lift is moving up and decelerating, then a<0 so T=mg+ma<mg
If lift is moving down and accelerating, then a<0 so T=mg+ma<mg
If lift is moving down and decelerating, then a>0 so T=mg+ma>mg
Tension in cable is less than weight if lit is moving up and decelerating or the lift is moving down and accelerating
Tension in cable is greater than weight if the lift is moving up and accelerating or lift is moving down and decelerating
Pulley problems
Resultant force on mass M, Mg-T+Ma
Resultant force on mass m, T-mg=ma
Combined:
Mg-mg=(M+m)a
Sliding down a slope:
mgsinθ-F0=ma
8.3 Terminal speed
Motion of an object falling in a fluid
Any object moving through a liquid will experience a drag force that depends on the shape of object, its speed and the viscosity of the fluid. The faster the object moves, the greater the drag force
The resultant force of an object moving through the fluid is the difference between the force of gravity on it(weight) and the drag forces on it
As drag forces increase(as it falls), resultant force decreases so the acceleration becomes less as it falls
When drag force is equal and opposite to the weight the object has reached
terminal velocity
At any instant, the resultant force F=mg-D
Therefore, acceleration of the object = mg-D/m=g-(D/m)
Motion of a powered vehicle
For a powered vehicle of mass m moving on a level surface, if FE represents
motive power
provided by the engine, the resultant force = FE-FR, where FR is the
resisitive force
opposing the motion of the vehicle
Therefore its acceleration
a=FE-FR/m
Maximum speed of the vehicle Vmax is reached when the resistive force becomes equal and opposite to the engine force(terminal velocity)
8.4 On the road
Stopping distances
Thinking distance
Distance travelled by vehicle in the time it takes the driver to react
For a vehicle moving at constant speed v, the thinking distance s1 =speed x reaction time=v x t0 where t0 is the reaction time of the driver(affected by distractions, drugs and alcohol)
Braking distance
Distance travelled by a car in the time it takes to stop safely, from when the brakes are first applied
Assuming constant deceleration, a, to zero speed from speed u, the braking distance s2 = u²/2a since u²=2as2
Stopping distance = thinking distance + braking distance = ut0 + u²/2a
Practical: Testing friction
Measure the limiting friction between the underside of a block and the surface it is on. Pull the block with and increasing force until it slides. The limiting frictional force on the block is equal to the pull force on the block just before sliding occurs> find out how this force depends on the weight of the block.
Vehicle Safety
Measuring impacts
The effect of a collision can be measured in terms of the acceleration and deceleration of the vehicle
By expressing an acceleration or deceleration in terms of g, the acceleration due to gravity, the force of the impact can then easily be related to the weight of the vehicle
Contact time and impact time
When objects collide and bounce off each other, they are in contact with each other for a certain time(same for both objects)
The shorter the contact time, the greater the impact force for the same initial velocities of the two objects
impact time t=2s/u+v
acceleration a=v-u/t
impact force F=ma
Car safety features
Vehicle bumpers
give way a little in a low-speed impact and so increase impact time
Crumple zones
are when the engine compartment of the car are designed to give way in a front-end impact(impact time increased)
Seat belts
are the wearers restraints when the vehicle suddenly stops. The seat belt stops the wearer more gradually than if they were not wearing one
Collapsible steering wheel
are steering wheels designed to collapse in the event of an impact to lessen the force
Airbags
act as cushion and increase the impact time on a person and impact area is increased