Book page 44-47 NETON S LAS OF MOTION
INERTIA Moving objects have inertia a property of all objects to resist a change in motion Mass: a measure of a body s inertia Two types of mass: - inertial mass m i - gravitational mass m g Since acceleration of free fall is independent of the mass of an object cgrahamphysics.com 2015 m i = m g eight: The gravitational force that Earth exerts on a body eight: varies with location, but inertial and gravitational mass stays constant
NETON S 1 ST LA If a body is at rest it wants to remain at rest and if the body is moving in a straight line with uniform motion it will continue to move with uniform motion unless acted upon by an external force. This does not take friction into account Friction increases with increasing speed
EXAMPLE OF THE FIRST LA hat are the forces acting on a ball thrown towards a partner? Aristotle view: Galileo view thrust
HAT DOES THE 1 ST LA TELL US? An object moving at constant velocity is in EQLB. All forces that act on it cancel out. Anything that is changing direction must have a resultant force acting on it accelerating, decelerating or changing direction Also called Law of Inertia b/c it has the tendency of matter described to keep moving in the way it is already moving
INERTIAL FRAME OF REFERENCE A system on which no forces act Laws of Physics are valid There is no experiment to distinguish between an observer at rest and one that is moving at constant velocity I am still Stand still
NETON S 2 ND LA If an unbalanced force acts on an object, the object accelerates the greater the mass the smaller the acceleration for a constant force a 1 m If the force doubles, acceleration doubles a F net Combining both observations: F net = ma
F net = ma Important: The net force is the sum of all forces acting on the object The acceleration is directly proportional to the force acting and is in the same direction as the applied force. If we express the acceleration in terms of rate of change of velocity, Newton s 2 nd law can be rewritten: F net = ma = m v t F net = p t another form of Newton s 2nd Law
EXAMPLE cgrahamphysics.com 2015 Newton s Laws hold true in an inertial frame a car going around a curve is an example of a non inertial frame In an extreme test on its braking system under ideal road conditions, a car traveling initial at 26.9ms 1 [S] comes to a stop in 2.61s. The mass of the car with the driver is 1.18 x 10 3 kg. Calculate a) the car s acceleration and b) the net force required to cause that acceleration Solution a) a = v u t = 0 26.9 2.61 = 10.3ms 2 b) F net = ma = 1.18 10 3 10.3 = 1.22 10 4 N[N]
EXAMPLE A man is riding in an elevator. The combined mass of the man and the elevator is 7.00 10 2 kg. Calculate the magnitude and direction of the elevator s acceleration, if the tension in the supporting cable is 7.50 10 3 N. T T ma + mg = T Assume up, if not sign will be negative a = T mg m = 7.50 103 7.00 10 2 10 7.00 10 2 = 0.71ms 2 up!
CONTINUED Find the tension in the cable if the acceleration is down T T = T + ma T = mg ma = m(g a) = 7.00 10 2 (10-0.71) = 6.5 10 3 N
NET FORCE IN FREE FALL The net force of an freely falling object is = mg But 2 nd law says F net = ma Hence mg = ma g = a This also shows the equivalence between inertial and gravitational mass Assume gravitational force of Earth on object is F g = km g where k is a constant The acceleration is given by Newton s 2 nd law: F g = km g = m i g where m i = inertial mass But experiments show that g is a constant and has the same value for all objects k = g and m g g = m i g m g = m i
DEFINITION OF FORCE 1N is the force which produces an acceleration of 1ms 2 in a mass of 1kg.
EXAMPLE Assume your weight is 55kg and you are standing on a scale in an elevator. If the scale is calibrated in Newton s, what is the reading on the scale when the elevator is not moving? = mg = 55 x 10 = 550 N If the elevator begins to accelerate upward at 0.74ms 2, what will be the reading on the scale? N ma ma N N = mg + ma = m(g + a) N = 55 (10 + 0.75) = 5.9 x10 2 N
EXAMPLE The diagram shows a block of wood of mass 1.0 kg attached via a pulley to a hanging weight of mass 0.5kg. Assuming that there is no friction between the block and the bench and taking g to be 10 ms 2, calculate the acceleration of the system and the tension in the string 1.0kg T ma T = 1a T T 0.5kg T + 0.5a = 0.5g a + 0.5a = 0.5g 1.5 a = 5 a = 5 1.5 = 3.3ms 2 T = 3.3N ma ma
EXAMPLE A person of mass 70 kg is strapped into the front seat of a car, which is travelling at a speed of 30ms 1. The car brakes and comes to rest after travelling a distance of 180m. Estimate the average force exerted on the person during the braking process. Solution v 2 = u 2 + 2as Minus sign slowing down a = v2 u 2 = 0 302 2s 2 180 = 2.5ms 2 F = ma = 70 x 2.5 = 175N