QUIZ 6 (REAL EXAM 2) |
ANSWERS |
SAMPLE TEST 2 |
QUIZ 6 IS MORE CLOSELY EXAMPLE DRIVEN---HERE WILL BE A CLOSE CORRESPONDENCE, often with explicit cross indexes, BETWEEN EXAMPLES 5.1 TO 5.24 AND THE EXERCISE/PROBLEMS. THESE PROBLEMS HELP US UNDERSTAND APPLICATIONS OF NEWTON'S LAWS, WHICH IN TURN OPEN US VIA ANALOGY TO DEEP INNER-WORKINGS OF NATURE. FOR EXAMPLE CLASSICAL MODELS STARTING WITH THESE PROBLEMS HELP US VISUALIZE MOLECULAR PROCESSES OBEYING NON-INTUITIVE STATISTICS. Sometimes it helps to think of an electron orbiting in a well defined, classical circular orbit around the nucleus using this chapter's definition of centripetal force even though we know electrons come in waves or "clouds" with a probability function and UNCERTAINTY PRINCIPLE . |
DISCUSSION QUESTIONS 12*, 13*, 16*, 19*, 20*, 23*, 24*, 25*, 26*, 27*, 28*, 29*, 30* |
EXERCISES/PROBLEMS: |
SEC. 5.1 (EQUILIBRIUM, STATIC OR DYNAMIC): 2*, 6( Try also 56,
8, 9) 34, SEE EXAMPLE 5.4, 60 (Try 59), 34; SEC 5.2 (DYNAMICS): #68*, A WARM UP PROBLEM, WHICH HAS A FAMILIAR GEOMETRY (CLASS NOTES) AND ILLUSTRATES NON-EQUILIBRIUM CONDITIONS; ALSO DO 12*, 14*, 16* , 17* (Try 15*) , 34* , 35*, 19*, 20* ; SEC. 5.3 (FRICTION AND AND/OR MORE COMPLICATED SITUATIONS): 26* , 29* (Try # 26*), 69* , 36* , 38* , 41* , 40* ; EXTENSIONS OF YOUR KNOWLEDGE BASE: (Try 97* and 122*, 123*) |
NOTE: #97* IS EXTREMELY IMPORTANT IMPORTANT FOR REASONS THAT CRYSTALLIZE AFTER TODAY. |
2*. You can also assume in each block is moving at constant
speed along the lines of their motion, which means the acceleration
equals zero. This is called DYNAMIC EQUILIBRIUM. In either case AT
REST OR WITH CONSTANT SPEED, with mass m, EACH
BLOCK has acceleration equal to zero , so you can simply write sum of forces in y dir = pos - nag = 0 since the acceleration in y direction = 0, In (a) for example ,we have 0 = pos - neg = T - mg. |
6*. Form an x-y axes on the plane of you page and let right
and upward on paper be the pos x and pos y directions,
respectively. ISOLATE MASS A. sum of force in x dir = 0 = -TA*cos0 + TB*cos theta, where theta is the angle the B string makes with the pos x-axis. sum of forces in y dir = 0 = TB*sin theta - mg |
8*. See #4, Ch. 4 to compare. DRAW YOUR AXES DIFFERENTLY NOW WITH an x-y axes on the plane
of your page and let UP the incline be the pos x direction and let
PERPENDICULAR and away from the incline surface be the
pos y direction. sum of forces in x direction = 0 = T*cos31 - mg*sin 25. sum of forces in y direction = 0 = T*sin 31 + N - mg*cos 25. From this info you can answer all questions selectively. DRAW A FREE BODY DIAGRAM. |
9*. . DYNAMIC EQUILIBRIUM MEANS MOTION WITH ZERO ACCELERATION AS IN
THIS CASE. DRAW YOUR AXES NOW WITH an x-y axes on the plane
of your page and let DOWN the INCLINE be the pos x direction and let
PERPENDICULAR and away from the INCLINE surface be the
pos y direction. (a) sum of forces in x direction = 0 = mg*sin 11 - F sum of forces in y direction = 0 = N - mg*cos 11 (b) sum of forces in x direction = 0 = mg*sin 11 - F*cos 11 sum of forces in y direction = 0 = N - mg*cos 11 - F*sin 11; Note in the increase in N from part (a) case. DRAW A FREE BODY DIAGRAM. |
SPECIAL PROBLEM 1(Test 2 review): FIND THE TENSION IN THE ROPE CONNECTING (a) BLOCKS A AND B (b) BLOCK A TO WALL. See Example 5.4 . To get the tension in the rope connecting A to the wall, you can assume that the that rope supports a total mass M_{A} + M_{B}. Thus T_{A} = (M_{A} + M_{B})*g*sin alpha, , where alpha is the angle with the horizontal. To find T_{B}, isolate M_{B}. : Let the x axis be along the incline so that sum of forces in x direction = 0 = T_{B} - M_{B}*g*sin alpha. For the normal forces in both cases we have: sum of forces in y dir = N = M*g*cos alpha |
60*.. DRAW YOUR AXES DIFFERENTLY NOW WITH an x-y axes on the plane
of your page and let UP the plane be the pos x direction and let
PERPENDICULAR and away from the plane surface be the
pos y direction. APPLY THE FORCES AT THE BALL CENTER. The tension exerts a force parallel to the horizontal , so it is like #9, part (b). DRAW A FREE BODY DIAGRAM. sum of the forces in x direction = 0 = pos - neg = T*cos35 - mg*sin 35 sum of forces in y direction = 0 = N - mg*cos 35 - T*sin 35 |
12*. See example 5.9. Let the power supply play the same role as
the woman. Note the difference. In this problem the rocket
moves up and accelerates upward ( speeds up ) whereas in the
example, the elevator slows down while moving down, which means the
acceleration also points up. So the acceleration is the same but the
motions are different between the example and this problem.
If up is positive for the motion, you'd write after isolating the
power supply: For part (b), m*a = pos - neg = N - mg, where m is the supply's mass and N is the magnitude of the normal force on it. Note: For part (a), (m + M)*a = F - (m + M)*g where F = 1720 (N) and M + m is the sum of the rocket's and POWER SUPPLY'S mass. |
14*.. See notes to capture the idea of two mass-less strings connecting 3 blocks together as they moved with common acceleration under the influence of an external force of magnitude F. Formally, in this problem you'll have three equations and three unknowns, acceleration a, and T_{A} and T_{B}. ISOLATE EACH MASS ! |
16*. (a) See class notes: Sum of the forces in the x dir = m*a = pos - neg = m*g*sin theta. DRAW YOUR AXES NOW WITH an x-y axes on the plane of your page and let DOWN the incline be the pos x direction and let PERPENDICULAR and away from the incline surface be the pos y direction. You can get acceleration a from Ch. 2 theory. (b) Add friction and the equation becomes: Sum of the forces in the x dir = m*a = pos - neg = m*g*sin theta. - f, where f = 10.0 (N). |
17*. See example 5.12. |
19*. Nice problem, harkening back to example
5.2 for
a string with mass.
Let upward be the +y direction. Let a lifting clamp labeled A
meet the top of the chain at point A. Label the
chain S just like I've done and the boulder B. Use
the same notation as notes but remember each object-- boulder B
and the chain S--have a downward gravitational force on them. For
example the string equation reads m_{s}*a = F_{SA} - F_{SB} - m_{s}*g. The boulder equation reads: m_{B}*a = F_{BS} - m_{B}*g. Note F_{SA} = the tension at the top of chain S = force of the clamp A on chain. Note: F_{BS} = the tension in string at the bottom. Note : F_{BS}= F_{SB } from Newton's 3rd Law. Now, if loading is done in dynamic equilibrium at constant speed directed upward, a = 0. Thus: F_{SA} = F_{SB} + m_{s}*g and F_{BS} = m_{B}*g. You can now find F_{SA} and F_{BS} . Note: Theoretically the tension at the midpoint of chain is (F_{SA} + F_{SB} )/2. For the problem at hand, let's consider the case when loading is not in equilibrium and upward acceleration a is greater than zero. Return to equations: m_{s}*a = F_{SA} - F_{SB} - m_{s}*g. m_{B}*a = F_{BS} - m_{B}*g. Set F_{SA} = 2.50*m_{s}*g. Solve for a. (b) Ch. 2. |
20*. The main thing is finding out if the magnitude of the
scale force reading is larger or smaller than the student's normal
weight mg. The scale force reading is equal to magnitude N of
the normal force on the student. If N >mg, the elevator
is accelerating upward. If N < mg, the elevator is accelerating
downward. Assume the motion and thus the positive direction
are upward without loss of generality. |
26*. See section 5.3 and figure 5.19's narrative. |
29.* SEE figure 5.19
narrative. (a) f_{sMAX} = 313 (N) = u_{s}*N; f_{k} = 208 (N) = u_{k}*N. (b) m*a = pos - neg = F - f_{k} , where a = 1.10 m/s^2. (c) Repeat if g = 1.62 m/s^2. |
69.*. Read page 151 about rolling
friction. Use CH. 2 methods friction
with constant acceleration, THIS TIME NEGATIVE, found from Newton's 2ND Law: m*a = - pos - neg, where pos = 0 in the case when the net horizontal force, pointing in the negative direction, is due to rolling friction only. |
36.* Examples 5.16 and 5.17. NOTE FOR PART (a), use the coefficient of static friction. |
38.*. Example 5.15. |
41.*. Example 5.18 (sec. 5.3); NOTE: TERMINAL SPEED IS PROPORTIONAL TO THE SQUARE ROOT OF YOUR WEIGHT, SO THE HEAVIER YOU ARE, THE LARGER YOUR FINAL SPEED (FOR THE SAME D) WHEN NET FORCE VANISHES; SEE EQUATION 5.13. |
40.* Let the positive y direction be up. Let a be the y
component of acceleration. In both cases set v = v_{term}/2,
where v_{term} is the terminal speed of section 5.3
The magnitude of the air resistance force is R= b*v^{2},
proportional to the speed squared. (a) sum of forces in y dir = m*a = pos - neg = 0 - mg - R, where R= b*v^{2}. (b) sum of forces in y dir = m*a = pos - neg = R - mg, , where R= b*v^{2}. . |