MASTERING PHYSICS CHAPTER 4

MASTERING PHYSICS CHAPTER 4 - 4.1. Solve: A force is...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a point of contact between the object and its environment. (iii) A very few forces, such as gravity and magnetism, are long-range forces that require no contact. (iv) A force is a vector quantity , having both a magnitude (or size) and a direction. (v) When multiple forces act on an object, the forces combine through vector addition to give a net force rr FF i net = .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4.2. Visualize: Assess: Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse walls.
Background image of page 2
4.3. Visualize:
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4.4. Model: Assume friction is negligible compared to other forces. Visualize:
Background image of page 4
4.5. Visualize: Please refer to Figure Ex4.5. Solve: Mass is defined to be m = 1 slope of the acceleration-versus-force graph A larger slope implies a smaller mass. We know m 2 = 0.20 kg, and we can find the other masses relative to m 2 by comparing their slopes. Thus m m mm 1 2 12 1 1 1 52 2 5 040 040 020 008 == = = = ⇒= = × = / / . .. slope 1 slope 2 slope 2 slope 1 kg kg Similarly, m m 3 2 32 1 1 1 25 5 2 250 250 020 050 = = = = × = / / . slope 3 slope 2 slope 2 slope 3 kg kg Assess: From the initial analysis of the slopes we had expected m 3 > m 2 and m 1 < m 2 . This is consistent with our numerical answers.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4.6. Model: An object’s acceleration is linearly proportional to the net force. Solve: (a) One rubber band produces a force F , two rubber bands produce a force 2 F , and so on. Because Fa and two rubber bands (force 2 F ) produce an acceleration of 1.2 m/s 2 , four rubber bands will produce an acceleration of 2.4 m/s 2 . (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2 m ). Using Fm a = , 2 F = (2 m ) a a = F / m = 0.6 m/s 2
Background image of page 6
4.7. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when the net force on an object is zero. Thus, it is incorrect to say that “force causes motion.” Instead, force causes acceleration . That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity that measures a change of motion. Newton’s second law quantifies this idea by stating that the net force r F net on an object of mass m causes the object to undergo an acceleration: r r a F m = net The acceleration vector and the net force vector must point in the same direction.
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4.8. Visualize: Solve: Newton’s second law is Fm a = . When F = 2 N, we have 2 N = (0.5 kg) a , hence a = 4 m/s 2 . After repeating this procedure at various points, the above graph is obtained.
Background image of page 8
4.9. Visualize: Please refer to Figure Ex4.9. Solve: Newton’s second law is Fm a = . We can read a force and an acceleration from the graph, and hence find the mass. Choosing the force F = 1 N gives us a = 4 m/s 2 . Newton’s second law yields m = 0.25 kg.
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4.10.
Background image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/11/2008 for the course PHYS 2211 taught by Professor Duncan during the Fall '08 term at Georgia Perimeter.

Page1 / 50

MASTERING PHYSICS CHAPTER 4 - 4.1. Solve: A force is...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online