Newton’s Laws of Motion I. Law of Inertia II. F=ma III. Action-Reaction

Newton’s Laws of Motion 1 st Law – An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force. 2 nd Law – Force equals mass times acceleration. 3 rd Law – For every action there is an equal and opposite reaction.

2 nd Law The net force of an object is equal to the product of its mass and acceleration.

Newton’s 2 nd Law proves that different masses accelerate to the earth at the same rate, but with different forces. We know that objects with different masses accelerate to the ground at the same rate. However, because of the 2 nd Law we know that they don’t hit the ground with the same force. F = ma 98 N = 10 kg x 9.8 m/s/s F = ma 9.8 N = 1 kg x 9.8 m/s/s

Motion on a Smooth Inclined Plane

Consider the forces acting on an object moving on an inclined plane of inclination  to the horizontal. The simplest case is one in which the object is sliding on the inclined plane under the action of the weight and normal reaction only.

The weight mg is resolved into two components. The component mg sin  is along the inclined plane. The component mg cos  is perpendicular to the inclined plane.

Motion takes place along the inclined plane. The resultant force on the object is mg sin  acting down-slope. By Newton’s Second Law of Motion, the acceleration a of the object also points down-slope. No matter the object is moving up-slope or down-slope, the resultant force and hence the acceleration points down-slope. F = ma mg sin  = ma a = g sin 

For object sliding on a smooth inclined plane The acceleration depends on the inclination of the plane only. It does not depend on the mass. Objects of different masses slide on the inclined plane with the same acceleration. The acceleration always points down-slope, independent of the direction of motion (velocity) of the object.

There is NO motion in the direction perpendicular to the inclined plane. By Newton’s First Law of Motion, the forces in this direction should balance each other. Hence the normal reaction R = mg cos  The normal reaction depends on the weight of the object and also it decreases with the inclination  The steeper the slope, the smaller is the normal reaction.

Example: A trolley of mass 0.5 kg is released from a point O on a smooth runway inclined at 30 o to the horizontal. Assume g = 10 m s -2 Find the time and the velocity of the trolley when it reaches a point A 2 m from O.

The acceleration a is uniform and independent of the mass of the trolley. a = g sin 30 o = 10  0.5 = 5 m s -2 The trolley reaches A with velocity v at time t. S = ut+0.5at 2 2 = 0  t   t 2 t = s v 2 = u 2 – 2aS v 2 = 0  + 2   2 v = 4.47m s -1 Solution: