# READ ALL CHECKPOINTS IN THE CHAPTER !
# STUDY THE NUMBERED OR UNNUMBERED FIGURES, GRAPHS, AND TABLES.
# IT DOES NOT HURT TO DO EXTRA END-OF-CHAPTER REVIEW QUESTIONS, PLUG AND CHUGS, RANKINGS , OR EXERCISES AND PROBLEMS.
#IF TIME PERMITS GET MORE PRACTICE WITH THE OUT OF SEQUENCE SAMPLE MULTIPLE CHOICE QUIZZES AT VERTICAL LEFT HAND COLUMN Q1, Q2, Q3, LINKS ETC HERE~ http://www.nvaphysics.com/P11AU11.htm .
|Quiz 2 (These problems will contribute to real quiz 2 next Thursday) Physics 11 Autumn 2011 --- Selected topics + exercises/problems from Ch. 2|
|Ranking Questions: 1, 2, 3 and 4.|
|Exercises: 12, 13, 14, 15, 20, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 38, 39, 40, 42|
|Problems 1, 2, 3, 4|
|1. Generally, the further way from the rope the smaller the tension; See page 27|
|2. Figure 2.6|
|3. a. The larger the mass, the greater the resistance to a change in state of motion. (b) In each case the upward support force is equal in magnitude to the downward weight. The mass is proportional to the weight , so the larger the mass, the larger the weight.|
| 4. (a) This a a trick question. There is no
friction force acting against the motion, so is any force even required
to keep the objects moving? I think not. The cause of the motion is
connected to the object's inertia & is sufficient to keep each
object moving at constant speed without human application of a
(b) Stopping each object in the same internal is a different story than part (a). The greater the mass and the associated property inertia, the more resistance to the stopping force. The greater the mass, the greater the stopping force needed to bring the object to rest in the same time interval.
|12. When you QUICKLY pull hard, the section of the paper near your hand at the front of the roll responds directly and simultaneously with your pull motion and with a large acceleration, and gathers speed quickly. The section of the paper away from your hand-- including the entire paper roll-- tends to remain at rest and does not react as quickly to the force as the front section because of the delay in the transmission of force to the other molecules much like sound travels in a medium--- air or a solid like steel. There are also inertial effects related to the mass and proportional inertia of the other molecules delaying their response to the force when it arrives. Thus the speed of other molecules away from the pull and originally at rest does not change as much as that of front molecules. The space between front and other molecules becomes so large the paper tears as it reaches its elastic limit .|
|13. In this case your body (torso) moves forward (accelerates )with the seat and your head tends to want to stay at its original speed since it does not feel any force or very little compared to the seat that's rigidly attached to the accelerating car. Thus the relative speed between torso and head causes the space between the two to increase rapidly, stretching your neck tissue and causing injury. See # 12.|
|14. See #13. The head rest causes the head and body to move together , avoiding the stretching of tissue|
|15. If you ignore the friction force between the bottom of your body and the seat, then your body tends to want to remain at its original forward speed whereas the the bus accelerates ( speeds up) in the forward direction (when driver presses on the accelerator pedal) or backward (slows down) during braking .|
|20. SEE FIGURE 2.6.|
|25. THIS IS CALLED DYNAMIC EQUILIBRIUM: MOVING WHEN THE NET FORCE IS ZERO.|
|26. See figure 2.6|
|27. The magnitudes of the tension in each parallel section of the rope must be equal and add up to her weight. The scale is only in one section of the string.|
|28. For the left image, see #27 as it pertains to two sections of string being in a parallel configuration and thus dividing the force into halves. For the right image, the section of the rope connected to Harry has a magnitude of upward tension equal to Harry's full weight. Since the net force on Harry is zero, the weight must be balanced by the upward tension but this time all the tension is in a single section of string !|
|31. In this case the downward weight must be balanced by an upward force discussed on page 28.|
|32. Consider the vertically and horizontally directed forces. In both the vertical and horizontal directions, the net force is zero. Since the book is at rest we know from # 31, the downward weight cancels the upward support force. Meanwhile, in the horizontal direction, if their is no push or pull force, the friction force is zero. On the other hand if you push or pull horizontally, then if the book remains at rest, the static friction force will point in the opposite direction with the same magnitude as the push or pull force.|
|33. READ APPENDIX D ON VECTORS AND THEIR COMPONENTS. See figure D.9 in appendix D, which deals with the components of an object's weight as it sits on an inclined plane. The component of your interest is labeled D in the diagram and is perpendicular to the surface. As stated in example 3 of appendix D, the support or "normal " force is not shown but does have the same magnitude and opposite direction as component D. Since weight component D decreases as the incline angle increases, so does the balancing support force D which keeps the object in equilibrium in the direction perpendicular to inclined surface.|
|34. Without loss of generality, assume the book rests on a horizontal surface. Friction acts horizontally and parallel to the surface NOT perpendicular to it. Pushing down on the book vertically does increase the upward, balancing support force which prevents the book from accelerating downward. Given the horizontal nature of friction, does friction come into play while pushing vertically down? What would be your answer if you pushed down at an angle other than 90 degrees with the horizontal? In that case you would have a force component parallel to the surface. Friction would act opposite this horizontal force component , whether nor not the book moves in that direction along table surface..|
|35. See #31 discussion comments.|
|36. See #31 and also figure 2.9|
|38. See figure 2.11.|
|39. See #38.|
|40. Let us consider the crate. Assume the
crate is in contact with the ground while the workman pull upward. The
sum of the forces in the vertical direction is zero since the crate is
in equilibrium. As the workmen pull upward, the support
force deceases as revealed in the following explanation: The
downward force on the crate is its weight. The total upward force is the
sum of the pull force of magnitude FW and the support force
of magnitude N. Since all these forces add to zero, we have the
following equation assuming upward is the positive vertical direction: 0
= FW + N - | weight |. If you solve for N, is it larger
or smaller than |weight |?
Meanwhile, let's examine what happens when the workmen completely lift the crate off the table. In that case both the crate weight and the workman's own weight would sum to a magnitude equal to that of the upward support force on his/her feet. Thus the workman would experience an increased support force on his/her feet. What does this tell you about how the support force on a workman's feet changes when the workmen lift the crate while it's still in contact with the floor?
|42. See figure 4.14 and related discussion. The acceleration is terminated when the upward air drag force reaches a certain value and we say the parachutist has reached his/her terminal speed. What is the magnitude of the air drag force at this moment?|
|CHECK OUT DISCUSSI0NS FOR CH. 3 WHEN THEY ARE POSTED.|