1. Forces Revision Newton’s First Law.
An object will remain at rest or continue to move with constant velocity unless acted on by an unbalanced force.
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2. Forces Revision Newton’s Second Law.
When an unbalanced force acts on an object it will accelerate. The rate of this acceleration is given by;
Where;
Fun = Unbalanced Force (N)
m = mass (kg)
a = acceleration (ms-2)
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3. Examples of Newton’s First Law;
A car being driven at constant velocity ~ forces balanced are the drive force of the car and friction.
A ship floating in the water ~ forces balanced are the weight of the ship and the upthrust caused by the water.
A parachutist who has reached “Terminal Velocity”.
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4. Examples of Newton’s Second Law;
A car moving off from rest – Drive Force >> Friction.
A ship sinking – Weight >> Upthrust
A parachutist who is slowing down after he/she opens the parachute. Air Resistance >> Weight
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5. More About Parachutists
The graph below shows how the velocity of a parachute jumper varies with time.
Velocity (ms-1)
Time (s)
Freefall
1st Terminal Velocity
Parachute Opens
2nd Terminal Velocity
lands
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6. Explanation of Graph
Freefall – Weight >> Air Resistance but Air Resistance increases the faster the parachutist falls.
1st Terminal Velocity - Weight = Air Resistance, no unbalanced force so constant velocity.
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7. Explanation of Graph
Parachute Opens – Air Resistance >> Weight but Air Resistance decreases as parachutist slows down.
Landing - Weight << Upwards Force due to ground so accelration to rest.
Note – We will learn why parachutists bend their knees on landing when we do Impulse.
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8. Definition of The Newton
One newton (N) is defined as the unbalanced force required to accelerate a mass of 1 kilogram (kg) with an acceleration of 1 metre per second per second (ms-2) 1N
Acceleration
= 1ms-2
1kg
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9. Example 1
A rocket of mass kg is required to lift off with an acceleration of 3ms-2 .
Find;
a) Find the size of thrust the engines must produce.
b) In reality will this acceleration increase, decrease or remain constant as the rocket climbs?
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10. Thrust Unbalanced Force = Thrust + Weight Mass = 10 000kg
weight = mg = x 9.8 = N
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11. Fun = Thrust + Weight
Also;
So we now have;
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12. This of course assumes a constant Thrust force from the engines.
In reality the rate of acceleration will increase as the mass of the rocket will decrease as it burns fuel.
The unbalanced force will therefore also increase as F = T – W and weight is getting smaller.
So we have a = F/m with F getting bigger and m getting smaller the higher it goes.
This of course assumes a constant Thrust force from the engines.
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13. Example 2
A gardener pulls a roller of mass 50kg at an angle of 45 with a force of N. If the force of friction on the roller is a constant 90N calculate the acceleration. N
90 N
45
Mass = 50kg
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14. cos 45 N
90N
Mass = 50kg
( cos 45) – 90 = 10N
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15. Example 3 – Accelerating Systems
Wile Coyote (mass 35kg) decides to lower, over a cliff, an anvil (mass 25kg) using a pulley wheel and rope onto Road Runners head. However Wile forgets to take off his frictionless roller blades and finds himself being accelerated towards the edge of the cliff as the anvil falls.
Find the tension force in the rope.
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16. Pulley
25kg
35kg
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17. The unbalanced force is provided by the weight of the anvil.
The system has a mass of 60kg (anvil + Wile).
System mass = (35kg (Wile) + 25kg (anvil)) = 60kg
Fun = W = mg (anvil) = 25 x 9.8 = -245N
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18. Firstly, calculate the acceleration of the system
Firstly, calculate the acceleration of the system. This is found by considering the TOTAL mass of the system and the accelerating force , gravitational force acting on the anvil.
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19. The tension is the force that causes Wile to accelerate at 4.08 m s-2.
To find the tension in the rope we need to consider the forces around a single part of the system, either will provide the answer. Consider Wile first;
Fun (tension)
Mass = 35kg
a = 4.08ms-2
The tension is the force that causes Wile to accelerate at 4.08 m s-2.
Fun = m x a = 35 x = N
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20. The same answer can be obtained by considering the anvil, this is slightly more complicated;
Tension
a = -4.08ms-2
Mass = 25kg
Weight = mg = 25x9.8 = -245N
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21. Example 4 – Choo Choo Type
A = 4kg
B = 3kg
C = 2kg
Driving Force = 27N 3N 3N 3N
Frictional force
Find the tension in the link between A and B.
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22. Step 1; we find the acceleration of the system.
9kg
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23. Step 2; consider the link between A and B
Tension
6N (Friction)
5kg
a = 2ms-2
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24. Example 5 – The Blocks
10kg
4kg
56N
Find the magnitude of the force exerted by the 4kg block on the 10kg block.
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25. Step 1; we find the acceleration of the system.
14kg
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26. Step 2; consider forces on the 10kg block
Fun
10kg
a = 4ms-2
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