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Chapter (5)

Chapter (5). Newton’s Laws of Motion. In this chapter we will study Newton's laws of motion.  These are some of the most fundamental and important principles in physics. Force. Intuitively, we know that force is a “push” or “pull”. Idea: Force is the cause of motion in classical mechanics.

abra-clements
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Chapter (5)

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  1. Chapter (5)

  2. Newton’s Laws of Motion In this chapter we will study Newton's laws of motion.  These are some of the most fundamental and important principles in physics.

  3. Force • Intuitively, we know that force is a “push” or “pull”. • Idea:Force is the cause of motion in classical mechanics. • Forces come in different classes (types): • Contact Forces : involve physical contact between objects Example: friction, viscosity etc… • Field Forces :don't involve physical contact between objects Examples: Gravity, Electromagnetism

  4. Force F is a vector quantity: You push or pull in a specific direction If force has direction, what is it’s measure? Fundamental force in nature

  5. In 1686, Newton presented his Three Laws of Motion: Newton’s First Law An object at rest remains at rest, and an object in motion continues in motion with constant velocity, unless it experiences a net force. • Velocity = constant (i.e. acceleration = 0) if there is no force (or if all forces add to zero). • Remember: • Velocity = constant, does not mean velocity = 0. • Velocity = constant means constant magnitude AND direction

  6. Newton's first law sometimes called the law of inertia Inertia Frames The tendency of an object to resist a change in its velocity is called inertia. The measure of inertia is mass. • SI units measure mass as multiples of the standard kilogram (kg=1000g) stored at the International Bureau of Weights and Measures in Sèvres, France. Newton’s First Law: If F=0, then a=0. What if F 0?

  7. Newton’s Second Law The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass. The direction of the acceleration is the direction of the resultant force.

  8. Force is a vector • The net force is the vector sum of all forces acting on the object m. • Mass is a scalar: • The value of the mass of an object does not change with the direction of the acceleration. • The equation F=ma is also a definition of mass. • Mass is invariant: • If two objects are put together (or separated) , the mass of the combined object is simply the arithmetic sum of the two masses m = m1+m2. The SI unit of force is One Newton is the force required to accelerate one kg one meter per second. Note that the first law is a special case of the second.  Is constant or zero

  9. The Vector Nature of Forces In the formula F = ma, F is the total (net) force acting on the object. We must consider the vector sum of all forces acting on an object. We can also consider each dimension separately:

  10. Example An object of mass 5 kg undergoes an acceleration of a = (8 m/s2) ŷ = 8 m/s2 in + y direction What is the force on that object? F = ma = (5 kg)(8 m/s2) ŷ = 40 kgm/s2ŷ ŷ = vector on unit length (no dimensions) in +y direction. The force is in the same direction as the acceleration.

  11. Example: Two forces, F1=45.0N and F2=25.0N act on a 5.00kg block sitting on a table as shown.  What is the horizontal acceleration (magnitude and direction) of the block? Solution: F1x= F1cos(65.0) = 19.0 N F2x= F2 = 25.0 N 19.0 - 25.0 = (5.00)ax ax = -1.2 m/s2

  12. Example: What is the average force exerted by a shot putter on a 7.0 kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s. Solution:

  13. Newton’s Third Law If object 1 exerts a force F on object 2, then object 2 exerts a force –F on object 1. • Forces come in pairs. • The force pairs act on different objects. • The forces have the same magnitude but opposite direction. Example: I push on the wall with a force of 20 N. The wall pushes back on me with a force of 20 N in the opposite direction.

  14. Weight The weight of any object on the Earth is the gravitational force exerted on it by the Earth: W = mg Note: Weight is a force (and therefore a vector). Weight is not equivalent to mass. The weight of an object is different on the earth and on the moon since the strength of the gravitational field is different ( g e g m ).

  15. Some application

  16. Definition of equilibrium If the net for force exerted on an object is zero, the acceleration of the object is zero and its velocity remains constant. That is, if the net force acting on the object is zero, the object either remains at rest or continues to move with constant velocity. When the velocity of an object is constant (including when the object is at rest), the object is said to be in equilibrium. The first condition of equilibrium

  17. Example: A traffic light weighing 122 N hangs from a cable tied to two other cables fastened to a support, as in Figure The upper cables make angles of 37.0 ° and 53.0 ° with the horizontal. Find the tension in the three cables. Solution: Note that the traffic light in equilibrium. Then

  18. since From the condition of equilibrium By solve the eqs. Example: Suppose the two angles are equal. What would be the relationship between T1 and T2?

  19. Example: A person weighs a fish of mass m on a spring scale attached to the ceiling of an elevator, as illustrated in Figure. Show that if the elevator accelerates either upward or downward, the spring scale gives a reading that is different from ward, the weight of the fish. Solution:

  20. We chose upward as the positive y direction (1) If the elevator is either at rest or moving at constant velocity, the fish does not accelerate, and so, Or (Remember that the scalar mg is the weight of the fish.) (2) If the elevator moves with an acceleration a relative to an observer standing outside the elevator in an inertial frame , Newton’s second law applied to the fish gives the net force on the fish: (*) Thus, we conclude from (1) that the scale reading T is greater than the weight mg if a is upward, so that ayis Positive, and that the reading is less than mg if a is downward, so that, ayis negative.

  21. For example: if the weight of the fish is 40.0 N and a upward, so that ward, ay = 2.00 m/ s2, the scale reading from (*) is (2) If a is downward so that ay= - 2.00 m/s2, then (*) gives us,

  22. Example: A car of mass m is on an icy driveway inclined at an angle as in Figure. Find the acceleration of the car, assuming that the, driveway is frictionless. Solution:

  23. Now we apply Newton Newton’s second law in component form, noting that ay= 0: (1) (2) From (1) From (2) Example: solve the example if the mass 1000 kg and the angle 30.

  24. Example: When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass, as in Figure, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to measure the free-fall acceleration. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord. Solution: When Newton’s second law is applied to object m1, we obtain

  25. Similarly , for object m2 we , find When (2) is added to (1), T cancels and we have Then we have Example: find the acceleration and the tension of Atwood's machine in which m1=2kg, m2=4kg.

  26. example: A ball of mass m1 and a block of mass m2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass, as in Figure. The block lies on frictionless incline of angle Find the magnitude of the acceleration of the two objects and the tension in the cord. Solution: Equations of motion for m1 (1) (2)

  27. Equations of motion for m2 (3) (4) Now, solve (2) , (3) simultaneously . The acceleration When this is substituted into (2) we find the normal force Example: if m1=10kg, m2=5kg and =45 find a and T.

  28. problem: Two blocks of mass 3.50 kg and 8.00 kg are connected by amassless string that passes over a frictionless pulley. The inclines are frictionless. Find (a) the magnitude of the acceleration of each block and (b) the tension in the string

  29. Example:  A 3.0 kg mass hangs at one end of a rope that is attached to a support on a railroad car. When the car accelerates to the right, the rope makes an angle of 4.0° with the vertical.  Find the acceleration of the car. Solution: In the vertical direction:                      

  30. In the horizontal direction: 

  31. Problem: A train is given an acceleration of 5 m/s2.  What is the tension between the two cars with a mass of 2000 kg.

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