Newton’s Laws of Motion. Newton’s Laws of Motion The First, Second and Third Laws Mass, Force, Inertial Frame Weight, the Normal Force, Friction Centripetal Force and Circular Motion. Newton’s First Law of Motion.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
This is Newton’s first law, which is often called the law of inertia:
Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.
Inertial reference frames:
Newton’s first law does not hold in every reference frame, such as a reference frame that is accelerating or rotating.
An inertial reference frame is one in which Newton’s first law is valid. This excludes rotating and accelerating frames.
How can we tell if we are in an inertial reference frame? By checking to see if Newton’s first law holds!
Mass is the measure of inertia of an object, sometimes understood as the quantity of matter in the object. In the SI system, mass is measured in kilograms.
Mass is notweight.
Mass is a property of an object. Weight is the force exerted on that object by gravity.
If you go to the Moon, whose gravitational acceleration is about 1/6 g, you will weigh much less. Your mass, however, will be the same.
Newton’s second law is the relation between acceleration and force. Acceleration is proportional to force and inversely proportional to mass.
Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.
Rocket propulsion can also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket.
Note that the rocket does not need anything to “push” against.
Weight is the force exerted on an object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight of an object of mass m is:
The force exerted perpendicular to a surface is called the normal force. It is exactly as large as needed to balance the force from the object.
A 65-kg woman descends in an elevator that briefly accelerates at 0.20g downward. She stands on a scale that reads in kg.
Two boxes connected by a cord.
Two boxes, A and B, are connected by a lightweight cord and are resting on a smooth table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force of 40.0 N is applied to the 10.0-kg box. Find (a) the acceleration of each box, and (b) the tension in the cord connecting the boxes.
A mover is trying to lift a piano (slowly) up to a second-story apartment. He is using a rope looped over two pulleys as shown. What force must he exert on the rope to slowly lift the piano’s 2000-N weight?
A small mass m hangs from a thin string and can swing like a pendulum. You attach it above the window of your car as shown. What angle does the string make (a) when the car accelerates at a constant a = 1.20 m/s2, and (b) when the car moves at constant velocity, v = 90 km/h?
A box of mass m is placed on a smooth incline that makes an angle θ with the horizontal. (a) Determine the normal force on the box. (b) Determine the box’s acceleration. (c) Evaluate for a mass m = 10 kg and an incline of θ = 30°.
Approximation of the frictional force:
Ffr = μkFN .
Here, FN is the normal force, and μk is the coefficient of friction, which is different for each pair of surfaces.
Our 10.0-kg mystery box rests on a horizontal floor. The coefficient of static friction is 0.40 and the coefficient of kinetic friction is 0.30. Determine the force of friction acting on the box if a horizontal external applied force is exerted on it of magnitude:
(a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.
A 10.0-kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is 0.30. Calculate the acceleration.
Will you exert less force if you push her or pull her? Assume the same angle θ in each case.
Two boxes and a pulley.
Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn’t stretch. As box B moves down, box A moves to the right.
This skier is descending a 30° slope, at constant speed. What can you say about the coefficient of kinetic friction?
Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. (a) If the coefficient of static friction is 0.40, determine what range of values for mass B will keep the system at rest. (b) If the coefficient of kinetic friction is 0.30, and mB = 10.0 kg, determine the acceleration of the system.
Motion in a circle of constant radius at constant speed.
Instantaneous velocity is always tangent to the circle.
This acceleration is called the centripetal, or radial, acceleration, and it points toward the center of the circle.
Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that
For an object to be in uniform circular motion, there must be a net force acting on it.
We already know the acceleration, so can immediately write the force:
We can see that the force must be inward by thinking about a ball on a string. Strings only pull; they never push.
There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome.
If the centripetal force vanishes, the object flies off at a tangent to the circle.
Force on revolving ball (horizontal).
Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.
Revolving ball (vertical circle).
A 0.150-kg ball on the end of a 1.10-m-long cord (negligible mass) is swung in a vertical circle. (a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. (b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the speed of part (a).
A small ball of mass m, suspended by a cord of length l, revolves in a circle of radius r = l sin θ, where θ is the angle the string makes with the vertical. (a) In what direction is the acceleration of the ball, and what causes the acceleration? (b) Calculate the speed and period (time required for one revolution) of the ball in terms of l, θ, g, and m.
Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed at which the entire centripetal force is supplied by the
horizontal component of the normal force, and no friction is required. This occurs when: