QUIZ 5 (ANSWERS) |
SAMPLE TEST 2 |
READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE. |
CH. 5 |
CH. 5 - Multiple choice (TBA) and exercise/problems equilibrium: 2 (try #15 ) , 3, 5( see #12), 6, 12, 14, 16 (see #29); accelerometers: 23, 25, 26; general no friction: 22, 27, 28 (try #29, done in class), 29*, 66, 29; friction: 34, 35, 36, 37, 39, 46, 47; drag forces: 51, 52; elastic forces (Hooke's Law) : 59, 60, 61. |
TURN IN :CH. 5~2*, 5, 6, 23, 26, 28, 39, 46, 52, 60 * *DONE IN CLASS. |
DISCUSSIONS PROVIDED. |
SELECTED DISCUSSIONS TO exercises/problems
BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. |
error log #1 --See corrections to #6 below regarding the
forces perpendicular to the incline. error log #2 --See corrections to #26 below regarding the leftward motion in which the car slows down; the cable would swing left in that case. |
5. SEE EXAMPLE 5.2 FOR A GEOMETRICAL 2D (2 dimensional)
system. Resolve the vector components along horizontal and
vertical axes.
(a) sum of forces in x direction = pos - neg = 0 = Tcos 10 - T*cos 10. The above equation provides insight but very little help in actually
finding T . So use the y-component equations: Solve for T. |
6. See example 5.3 for a simplified version of problem; This time, in
contrast to the previous problem (#5), we resolve our components along x
and y axes that are parallel and perpendicular to the inclined plane, respectively.
Let the positive x direction be UP the incline. SUM of forces in x direction = pos - neg = 0 = T*cos 31 - mg*sin 25. SUM of forces in y direction = pos - neg = 0 = N + T*sin31 - mg*cos 25. Use these equations to answer parts (b) and (c) to find T AND N; use the definition of normal force, exerted by the surface on the car. |
23. See example 5.5 for the spirit of
this problem. Let a_{y} be the y-component of acceleration . (a) Let up be the positive y direction, the assumed direction of motion. sum of forces in y direction = m*a_{y }= pos - neg = F_{s} - mg, where F_{s} is the magnitude of the spring force on the fish; note the value F_{s} is the scale reading ! The scale reading is F_{s }= 60.0 (N). Now, you are given a_{y }= + 2.45 m/s^{2}. Solve for m and find mg. (b) Let DOWN be the positive y direction, the assumed direction of motion. In the previous part, we we see that mg < 60.0 (N). This must be true otherwise the fish would not be speeding up as it moves upward. If the fish accelerates downward, i.e., speeds up as it moves down, we have a different situation. In that case, mg > Fs. Assuming the object moves down, we have sum of forces in y direction = m*a_{y }= pos - neg = mg - F_{s }. You are given F_{s } = 35.0 (N). Solve for the acceleration component. (c) ZERO in which case the fish would be in free fall. |
26. See example 5.5 for a direct
application of this problem to a real world situation. As the car moves
right, speeding up, the acceleration points right and the cable swings left.
Meanwhile, as the car moves left, the cable would swing right if
the speed was increasing. But the speed decreases during leftward
motion, which means the acceleration points right and thus the cable
swings left as in part (a) . Note in part (c) the cable will hang vertically with no deflection since the acceleration is ZERO. |
28. See class notes on #29 we worked
on the board. Let a
be the magnitude of the common acceleration of the boxes.
Isolate the 373-N mass: Let this mass be labeled M. |
39. (a) sum of forces in x direction = pos
= neg = 0 since the velocity is constant as the box moves along the
horizontal. 0 = F - f_{k}, where F is the magnitude of person's push force and f_{k} is the friction force magnitude. Note: = u_{k}*N, where u_{k} is the coefficient of kinetic friction and N is the normal force magnitude. sum of the forces in the y direction = pos - neg = 0 = N - mg;
thus N = mg. From part (a), N = mg as before, but now we have a different situation in the x-direction: sum of forces in x direction = m*a_{x} = pos - neg = 0 - f_{k}, since there is no positively directed force ! Find acceleration a_{x} , which will be negative. |
46. (a) Let up the incline be the positive
x direction.
Going up the incline, sum of forces in the x direction = m* a_{x}
= pos - neg = 0 - f_{k} - mg*sin 40 since all the
x-directed -force components point down incline; there is no
"pos" force along x axis ! Note: = u_{k}*N,
where u_{k} is the coefficient of kinetic friction and N is the normal
force magnitude. |
52. See example 5.13: Let down be positive; sum of forces in y direction = m*a = mg - F_{D}, where a is the magnitude of the downward acceleration and F_{D} is the drag force magnitude. In the formula for F_{D} set v_{y} = v_{T }/ 2 , where the terminal velocity v_{T} is derived in the example. |