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Forces

Forces. Forces - LOs. (a) Solve problems using the relationship: net force = mass x acceleration ( F = ma ) appreciating that acceleration and the net force are always in the same direction; (b) define the newton ;

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Forces

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  1. Forces

  2. Forces - LOs • (a) Solve problems using the relationship: • net force = mass x acceleration (F = ma) appreciating that acceleration and the net • force are always in the same direction; • (b) define the newton; • (c) apply the equations for constant acceleration and F = ma to analyse the motion of objects; • (d) recall that according to the special theory of relativity, F = ma cannot be used for a particle travelling at very high speeds because its mass increases.

  3. Newton’s Laws of motion

  4. Newton’s First Law An object stays still or keeps moving in a straight line at the same velocity unless a resultant force makes it change. For speeding up, slowing down or changing a direction there is always a force involved.

  5. Newton’s Second Law Acceleration of an object is directly proportional to the resultant force exerted on it and inversely proportional to its mass. F a = m The harder you push something the faster it will speed up, and the smaller the object is, the faster it will speed up.

  6. Newton’s Third Law For every action there is an equal and opposite reaction. If you jump in the air. As you push your feet against the ground, the ground pushes back and causes you to go up to a height of, say 1 meter. However, your feet pushing against the ground also cause the Earth to move in the opposite direction from you. Because the Earth is a lot bigger and heavier than you, it only moves a tiny bit. ( about 0.00000000000000000000001 meter)

  7. The Newton • One Newton is the force that causes a mass of one kilogram to have an acceleration of one metre pre second every second

  8. Relativity • Einstein’s theory of special relativity says that as an object starts travelling very fast (near to the speed of light), it’s mass increases. Therefore we can’t use F=ma for these objects

  9. Summary Q • A car of mass 800 kg accelerates uniformly along a straight line from rest to a speed of 12 ms-1 in 50 seconds. Calculate: • The acceleration of the car • The force on the car that produced this acceleration • The ratio of the accelerating force to the weight of the car. • The acceleration of the car ‘a’ • a = (v-u)/t = (12 – 0)/50 = 0.24 ms-2 b. The force on the car that produced this acceleration ‘F’ F = ma F = 800 x 0.24 = 192 N c. The ratio of the accelerating force to the weight of the car. F/mg = 192/(800x9.8) = 0.024

  10. Summary Q • 2. An aeroplane of mass 5000 kg lands on a runway at a speed of 60ms-1 and stops 25 seconds later. Calculate: • The deceleration of the aeroplane • The braking force on it • The acceleration of the aeroplane ‘a’ • a = (v-u)/t = (0 –60)/25 = -2.4 ms-2 b. The braking force ‘F’ F = ma F = 5000 x -2.4 = -12000N The negative sign indicates the direction of the opposing force.

  11. 3. A vehicle of mass 1200 kg on a level road accelerates from rest to a speed of 6.0 ms-1 in 20 s, without change of direction. • Calculate the force that accelerates the car. • The vehicle now is fitted with a trailer of mass 200 kg. Calculate the time taken to reach a speed of 6.0ms-1 from rest for the same force in part a. • Acceleration = (v-u)/t = (6 – 0)/20 = 0.3ms-2 • F = ma = 1200 x 0.3 = 360 N • Total mass = 1400 kg • Force = 360 N • a = F/m = 360/1400 = 0.26 ms-2 • a = (v-u)/t • t = (v-u)/a = (6 – 0)/0.26 = 23 s

  12. 4. A bullet of mass 0.002 kg travelling at a speed of 120 ms-1 hit a tree and penetrated a distance of 55 mm into the tree. Calculate: • The deceleration of the bullet • The impact force of the bullet on the tree. • The deceleration of the bullet • v2 = u2 + 2as • 0 = (120)2 + 2 x a x 55x10-3 • -(120)2 = 2 x a x 55x10-3 • a = -(120)2 / (2 x 55x10-3) = -130909 ms-2 b. The impact force of the bullet on the tree. F = ma = 0.002 x -130909 = -261.8 N The negative sign indicates that the direction of the force is in opposite direction of the movement of the bullet.

  13. A car of mass 1200 kg has an engine which provides an engine force of 600N. Calculate: • Its initial acceleration • Its acceleration when the drag force is 400N • Acceleration = Force/mass • a = 600/1200 = 0.5ms-2 • b. When the drag force is 400N, the resultant force is equal to the weight – drag • resultant force = 600 – 400 = 200N • a = 200/1200 = 0.16kg

  14. A rocket of mass 550 kg blasts vertically from the launch pad at an acceleration of 4.2 ms-2. Calculate: • the weight of the rocket • The thrust of the rocket engines Thrust T • the weight of the rocket ‘W’ • W = mg • W = 550 x 9.8 = 5390 N b. The thrust of the rocket engines ‘T’ The resultant force F = ma F = 550 x 4.2 = 2310 N F = T – W T = F + W = 2310 + 5390 = 7700 N Weight W

  15. A car of mass 1400 kg pulling a trailer of mass 400 kg accelerates from rest to a speed of 9.0 ms-1 in a time of 60 s of a level road. Calculate a. The tension in the two bar b. The engine force • The tension in the two bar • Acceleration a = (v-u)/t = (9-0)/60 = 0.15 ms-2 • F = ma = 400 x 0.15 = 60N b. The engine force = the force to accelerate the car + the force to accelerate the trailer The engine force = (1400 x 0.15) + (400 x 0.15) = 270 N

  16. A lift and its occupants have a total mass of 1200 kg. Calculate the tension in the lift cable when the lift is: • Stationary • Ascending at a constant speed • Ascending at a constant acceleration of 0.4 ms-2 • Descending at a constant deceleration of 0.4 ms-2 • Stationary: T = weight of the lift = 1200 x 9.8 = 11760N • Ascending at a constant speed: same as when stationary • Ascending at a constant acceleration of 0.4 ms-2: • Force causing the acceleration = ma = 1200 x 0.4 = 480N • Total tension in the lift = 11760 + 480 = 12240 N • d. Descending at a constant deceleration of 0.4 ms-2 • Force causing the deceleration = ma = 1200 x 0.4 = 480N • Total tension in the lift = 11760 + 480 = 12240 N

  17. A brick of mass 3.2 kg on a sloping flat roof, at 30° to the horizontal slides at constant acceleration 2.0 m down the roof in 2.0 seconds from rest. Calculate: • The acceleration of the brick • The frictional force on the brick due to the roof • The acceleration of the brick • s = ut + ½ a t2 • 2 = 0 + ½ a 22 • a = 1 ms-2 N F mg sin30° mg cos30° b. The frictional force on the brick due to the roof: mg 30° Since the brick is accelerating there must be a resultant force on it in the direction of the roof. Resultant force = ma = 3.2 x 1 = 3.2 N Resultant force = mgsin30 - Friction 3.2 = (3.2x9.8x0.5) - Friction Friction = 12.48 N

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