1. Instructor: Sujood Alazzam
PHYS 101
General Physics
Newton’s laws of Motion
Work, Energy and Power
Fluids
Direct Current (DC)
Nerve Conduction
Ionizing Radiation
Wave properties of light
Instructor: Sujood Alazzam
2017/2018

2. CHAPTER OUTLINE Newton’s laws of Motion:
3.1 Force, Weight, and Gravitational Mass.
3.3 Newton’s first law.
3.4 Equilibrium
3.5 Newton’s third law
3.6 Newton’s second law
3.10 Weight.
3.12 friction.
9.2 The weight on a spring.
9.3 The physical pendulum.

3. Lecture 3: Using Newton’s Laws

4. Objectives
Demonstrate an understanding of the meaning of net force and be able to calculate the acceleration that results.
Define and give examples of the “normal force,” “friction force,” and “tension force” ; use conventional variables to represent these forces.
Distinguish among mass, and weight, calculate the weight of an object on the earth’s surface.

5. Using Newton’s Laws 3.10 Weight
Weight is a force caused (on Earth) by the gravitational attraction of a mass to the Earth’s centre. The weight of a body, of mass m, is defined to be the force, W, with which it is attracted to the Earth. On Earth, W = mg, where g is the acceleration due to gravity (g ≈ 9.81 m s−2 on Earth). FW = W = m g

6. Tension
Many mechanics problems involve objects being pulled, pushed or suspended from a string, spring, rod or something similar.
The force that the string (or similar) exerts on the object in these types of problems is called tension. T

7. Normal reaction force FN = N
A mass lies on a horizontal surface. The weight of the mass pulls it downwards. The reason it does not fall is because the horizontal surface exerts an equal and opposite force on the mass called the normal reaction force. The normal reaction force always acts perpendicularly to the surface that is causing it.
FN = N

8. Normal force
Fnet  0

10. 3.12 Friction
Friction is a force that always acts to resist the motion of one object sliding on another. Frictional forces are very important, since they make it possible for us to walk, and hold books.
When we walk or run, we are not conscious of any friction in our knees or other leg joints. These and many other mammalian joints are well lubricated by synovial fluid, which is squeezed through the cartilage lining the joints when they move (Fig ). This lubricant tends to be absorbed when the joint is stationary, increasing the friction and making it easier to maintain a fixed position.

11. Application of Friction
Kinetic Friction
Static Friction

12. Now suppose we apply a small horizontal force T to the right (Fig. 3
Now suppose we apply a small horizontal force T to the right (Fig. 3.19c). If the block remains at rest, the friction force fs can no longer be zero, since the first law requires that the net force be zero, or fs = T. If T is gradually increased, fs increases also. Eventually when T becomes large enough, the block begins to slide. Thus there is a maximum possible static friction force fs(max).

13. Experimentally, it is found that Fs(max) has the following properties:
Fs(max) is independent of the contact area
For a given pair of surfaces, Fs(max) is proportional to the normal force N.
The number relating Fs(max) and N, called the coefficient of static friction, μs , is defined by
The coefficient μs depends on the nature of the two surfaces, and on their cleanliness and smoothness.
Fs(max) = μs N

14. μk is nearly independent of the velocity, and, since. Fk  Fs μk  μs
The force necessary to keep an object sliding at constant velocity is smaller than that required to start it moving. Thus kinetic friction force fk is less than fs( max). It is independent of the contact area, and it satisfies:
Fk = μk N
Here μk is the coefficient of kinetic friction and is determined by the nature of the two surfaces.
μk is nearly independent of the velocity, and, since.
Fk  Fs
μk  μs

15. Example A 50 N block is on a flat, horizontal surface (Fig. 3.19).
(a) If a horizontal force T = 20 N is applied and
the block remains at rest, what is the frictional force?
(a) Since the block remains at rest when the force T is applied, the frictional force fs must be equal but opposite to T. Consequently fs = T = 20 N

16. (b) The block starts to slide when T is increased to 40 N. What is μs ?
Since the block just begins to slide when the applied force is increased to 40 N, the maximum frictional force must be fs(max) = 40 N
The vertical forces must add to zero, so the normal force N is equal to the weight, w, or 50 N. Hence
(c) The block continues to move at constant velocity if T is reduced to 32 N. What is μk?
Since the block moves with constant velocity when a 32 N force is applied, the net force must be zero. Consequently the frictional force Fk must equal the applied force, or Fk = 32 N

17. We now give several examples of how Newton's laws of motion are used.

18. In each case, we employ a systematic procedure for relating the acceleration of an object or objects to the forces present.
For each object, we draw a careful sketch showing the forces acting on that object.
We then apply Newton's second law F = ma to each object separately. If there are n forces F1 , F2 , F3 acting on an object, the net force F is the sum of the forces, and we have
F = F1 + F2 + • • • + FN = ma
In component form, this is
F = F1X + F2X + • • • + FN = maX
F = F1Y + F2Y + • • • + FN = maY
we solve the equations for the unknown quantities symbolically and then substitute the numbers if they are given.
We include the units of the quantities and see that the final answer has the correct dimensions.

19. Example
An elevator has a mass of 1000 kg. It accelerates upward at 3 m/s2. What is the force T exerted by the cable on the elevator?

20. The forces on the elevator are its weight w and the upward force T resulting from the cable
Note that T is greater than the weight mg. The cable must support the weight of the elevator and also provide the extra force needed for the acceleration.

21. Example
A child pulls a train of two cars with a horizontal force F of 10 N. Car 1 has a mass m1= 3 kg. and car 2 has a mass m2 = 1 kg (Fig. 3.11). The mass of the string connecting the cars is small enough so it can be set equal to zero, and friction can be neglected,
Find the normal forces exerted on each car by the floor,
What is the tension in the string?
What is the acceleration of the train?

22. In this problem, there is a system of several objects
In this problem, there is a system of several objects. The three objects of interest are the two cars and the string connecting them. When Newton's second law F = ma is applied to each object, a set of equations results that must be solved simultaneously. The fact that the string is massless means that the net force F = ma on the string must be zero. As we have explained, this implies that the forces it exerts on the two cars are equal in magnitude, so we have labeled its tension force by the same symbol T in the force diagrams for each of the cars.

25. Example
A block of mass m1 = 20 kg is free to move on a horizontal surface. A rope, which passes over a pulley, attaches it to a hanging block of mass m2 = 10 kg (Fig. 3.12). Assuming for simplicity that the pulley and rope masses are negligible and that there is no friction, find
(a) the forces on the blocks and (b) their acceleration?

26. Here one mass moves horizontally and one moves vertically
Here one mass moves horizontally and one moves vertically. The force exerted by the rope on m1 is equal in magnitude to the force exerted by the rope on m2 Accordingly we denote both forces by the same symbol T.
We first apply F = ma to the block on the surface. Since it has no vertical acceleration component, the net vertical force component must be zero. Thus the normal force N, on block 1 due to the surface is
N1 = W1, = m1g = (20 kg)(9.8 m/s2 )= 196 N
T = m1 a

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