Presentation is loading. Please wait.

Presentation is loading. Please wait.

As this windsurfer is propelled through the air, his motion is determined by forces due to the wind and his weight. The relationship between the forces.

Similar presentations


Presentation on theme: "As this windsurfer is propelled through the air, his motion is determined by forces due to the wind and his weight. The relationship between the forces."— Presentation transcript:

1

2 As this windsurfer is propelled through the air, his motion is determined by forces due to the wind and his weight. The relationship between the forces acting on an object and the resulting motion is discussed in this chapter. (Tom King/The Image Bank/Getty Images)

3

4  Definition: An inertial reference frame is one in which Newton’s law of inertia is valid.  The acceleration of an inertial reference frame is zero, so it moves with a constant velocity. All of Newton’s laws of motion are valid in inertial reference frames, and when we apply these laws, we will be assuming such a reference frame. In particular, the earth itself is a good approximation of an inertial reference frame.

5  When a net external force acts on an object of mass m, the acceleration a that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force.  SI Unit of Force: kg·m/s 2 = newton (N)

6 Two people are pushing a stalled car, (figure). The mass of the car is 1850 kg. One person applies a force of 275 N to the car, while the other applies a force of 395 N. Both forces act in the same direction. A third force of 560 N also acts on the car, but in a direction opposite to that in which the people are pushing. This force arises because of friction and the extent to which the pavement opposes the motion of the tires. Find the acceleration of the car.

7 A man is stranded on a raft (mass of man and raft = 1300 kg). By paddling, he causes an average force of P = 17 N to be applied to the raft in a direction due east (the +x direction). The wind also exerts a force on the raft that has a magnitude of A = 15 N and points 67° north of east. Ignoring any resistance from the water, find the x and y components of the raft’s acceleration.

8 Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body.

9 Fundamental 1. Gravitational force. 2. Strong nuclear force. 3. Electroweak force. are the ones that are truly unique, in the sense that all other forces can be explained in terms of them. Non-fundamental 1. The normal force. 2. Static and kinetic frictional forces. 3. The tension force.

10  Every particle in the universe exerts an attractive force on every other particle. A particle is a piece of matter, small enough in size to be regarded as a mathematical point. For two particles that have masses m 1 and m 2 and are separated by a distance r, the force that each exerts on the other is directed along the line joining the particles and has a magnitude given by The symbol G denotes the universal gravitational constant, whose value is found experimentally to be G = 6.673 x 10 -11 N.m 2 /kg 2

11 What is the magnitude of the gravitational force that acts on each particle in the above figure, assuming m 1 = 12 kg, m 2 = 25 kg, and r = 1.2 m? (You exert a force of about 1 N when pushing a doorbell, so F is exceedingly small in such circumstances). This is due to the fact that G itself is very small. However, for large masses, like that of the earth, the gravitational force can be large.

12  Definition: The weight of an object on or above the earth is the gravitational force that the earth exerts on the object.  The weight always acts downward, toward the center of the earth.  On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. SI Unit of Weight: newton (N)

13 Definition of the Normal Force  The normal force F N is one component of the force that a surface exerts on an object with which it is in contact.  namely, the component that is perpendicular to the surface.

14

15 Therefore, the seventh cervical vertebra exerts a normal force of F N = 50 N.  Figure (c) shows the free-body diagram that applies during the act. Now, the total downward force exerted on the standing performer’s head and neck is 490 N + 50 N = 540 N, which must be balanced by the upward normal force, so that F N = 540 N. Figure (b) shows the free- body diagram for the standing performer’s head and neck before the act. The only forces acting are the normal force and the 50-N weight.

16 (a) When the elevator is not accelerating, the scale registers the true weight (W = 700 N) of the person. (b) When the elevator accelerates upward, the apparent weight (1000 N) exceeds the true weight. (c) When the elevator accelerates downward, the apparent weight (400 N) is less than the true weight. (d) The apparent weight is zero if the elevator falls freely—that is, if it falls with the acceleration due to gravity.

17 Definition of Equilibrium: An object is in equilibrium when it has zero acceleration. When an object is accelerating, it is not in equilibrium. i.e. the net force is not zero in Newton’s second law

18  1/451) An airplane has a mass of 3.1 x 10 4 kg and takes off under the influence of a constant net force of 3.7 x 10 4 N. What is the net force that acts on the plane’s 78-kg pilot?

19  2/451) A boat has a mass of 6800 kg. Its engines generate a drive force of 4100 N, due west, while the wind exerts a force of 800 N, due east, and the water exerts a resistive force of 1200 N due east. What is the magnitude and direction of the boat’s acceleration?

20  3/451) In the amusement park ride known as Magic Mountain Superman, powerful magnets accelerate a car and its riders from rest to 45 m/s (about 100 mi/h) in a time of 7.0 s. The mass of the car and riders is 5.5 x 10 3 kg. Find the average net force exerted on the car and riders by the magnets. Solution: v = v o + a tF = m a 45 = 0 + (a) (7.0 s)F = 5.5 x 10 3 x 6.42 a = 6.42 m/s 2 F = 3.51 x 10 4 N

21  5/451)A 15-g bullet is fired from a rifle. It takes 2.50 X 10 -3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

22 11/452) Two forces, F 1 and F 2, act on the 7.00- kg block shown in the drawing. The magnitudes of the forces are F 1 = 59.0 N and F 2 = 33.0 N. What is the horizontal acceleration (magnitude and direction) of the block?

23 12/452) When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 N and the drag force has a magnitude of 1027 N. The mass of the sky diver is 93.4 kg. What are the magnitude and direction of his acceleration?


Download ppt "As this windsurfer is propelled through the air, his motion is determined by forces due to the wind and his weight. The relationship between the forces."

Similar presentations


Ads by Google