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HOMEWORK SET #1
HOMEWORK SET #2
| Problem |
Hint |
Answers |
| 3.1 |
When the car loses contact
the normal force goes to zero. |
r = 668 ft, N = 120 lbf |
| 3.2 |
Remember the normal force is
zero when the block leaves the incline. |
q = 27.5, x = 3.81 ft |
| 3.3 |
|
v = 7.83 ft/s, N = 1.45 lbf |
| 4.1 |
|
vr = -15.25
in/s, vq
= 32.52 in/s
ar = -128.8 in/s2, aq=
9.4 in/s2 |
| 4.2 |
Remember you know the
direction the plane is traveling. |
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| 4.3 |
To find rdot,
you need to separate variables and integrate. When solving the
problem you actually get rdot to be plus or minus 4.94 ft/s (because of
the square root). The answer to the right assumes we want the
velocity the first time the mass passes r = 0.5 ft. |
N = 2.30 lbf
vr = -4.94 ft/s, vq
= 7.5 ft/s |
HOMEWORK SET #3
| Problem |
Hint |
Answers |
| 6.1 |
|
vAfinal
= 1.013 m/s
vBfinal = 0.338 m/s
vCfinal = 0.15 m/s |
| 6.2 |
The spring force
is not impulsive. |
h = 0.072 ft
k = 72.2 lbf/ft |
| 6.3 |
|
e = 0.923
h1 = 1.28 m |
| 7.1 |
|
vB = 2.74 m/s at 42.1 deg CCW from horizontal |
| 7.2 |
|
q = 62.7 degrees
Elost = 0.14 vA2 |
HOMEWORK SET #4
| Problem |
Hint |
Answers |
| 10.1 |
|
a) q = 165
rev
b) q = 2200 rev |
| 10.2 |
If you first solve for q as a
function of time you should get q = 0.8t-0.9t2. |
vA = 1.64 m/s (up)
xA = 3.7 m (up)
vB = 0.738 m/s (down)
xB = 1.665 m (down) |
| 10.3 |
Rather than finding the
center of gravity for the composite shape it is often easier just to put the weights (for
the FBD) and the maGs (for the KD) at the c.g. for each individual bars. |
a = 20.7 ft/s2
Tc = -4.05 lbf
Tb = -14.3 lbf |
| 11.1 |
The disks will stop slipping
when the velocity of the point of contact is the same for both disks. Be sure to write the
velocities as functions of time and then solve for the time. |
a) aA
= 12.5 rad/s2 CCW
aB = 33.3 rad/s2 CCW
b) wA = 240 rpm CW
wB = 320 rpm CCW |
| 11.2 |
Be sure your answers include
the directions for Cx and Cy. |
a) a = 43.6
rad/s2 CW
b) Cx = 21 N, Cy = 54.6 N |
HOMEWORK SET #5
| Problem |
Hint |
Answers |
| 12.1 |
The spring is stretched at
both positions. |
w = 11.1 rad/s CCW |
| 12.2 |
The
answers shown for part c) neglect the mass of the bullet since we don't
know exactly where it stops in the plate and its mass is much less than
the mass of the plate. |
a) vG
= 2.28 m/s (to the right)
b) Ax = 1.26 kN (to the right)
Ay = 0
c) Ax = 361.9 N (to the left)
Ay = 35.3 N (up) |
| 13.1 |
For
part c) be sure to consider that the center of gravity is moving and it is
rolling. |
a) w = 3
rad/s CW
b) vA = 9 in/s to the right
c) 15 in of cord unwound per second |
| 13.2 |
To find the velocity of the
midpoint I think it is easier to use the vector algebra approach rather than instantaneous
center of velocity. |
a) wBD
= 12 rad/s CCW
b) 3.9 m/s at 67.4 |
| 14.1 |
Find w2
as a function of b and differentiate. |
b=L/sqrt(12)
w
= 1.86sqrt(g/L)
Cy=2mg |
| 14.2 |
The hardest part
in this problem is to relate all the angular velocities and velocities of the centers of
gravities in the kinetic energy expression. You will also need to use the velocity
of point B. Also, don't forget the work done by the moment. |
vGrod
= 0.48 m/s |
HOMEWORK SET #6
| Problem |
Hint |
Answers |
| 15.1 |
Be
careful when finding the work done by the force P. What distance
does it travel? |
vB =
16.3 ft/s |
| 15.2 |
|
vGgear
= 0.829 m/s (to the left) |
| 16.1 |
This
answer assumes the force the cue applies to the ball is horizontal. |
2/5r |
| 16.2 |
|
8.8
ft/s |
HOMEWORK SET #7
| Problem |
Hint |
Answers |
| 19.1 |
You
may assume the mass moments of inertia of the rods about their own CGs are
zero when they are vertical since you are not given the radius of the
rods. |
32.0 rev/min |
| 19.2 |
The velocity of
the point of contact between the sphere and the cart is not zero. Therefore, the point
of contact is not the instantaneous center of velocity. Therefore, I would suggest
using the vector algebra approach to relate vcart to vGsphere and wcyl.
You will need to use conservation of energy plus another conservation principle in order
to get enough equations to solve the problem. Since you are not given the radius of
the sphere, r, just leave it in your equations (note: it is not one of your
unknowns). When solving the equations if you get a RootOf just use
evalf(allvalues(solve( ...))) |
vcart
= 2.10 ft/s |
| 20.1 |
|
a) x = 0.6
m from A
b) x = 0.2 m from A |
| 20.2 |
|
a) aB
= 5407 ft/s2 (down)
b) aC = 5407 ft/s2 (up)
c) aD = 5407 ft/s2 (down at 60) |
| 20.3 |
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HOMEWORK SET #8
| Problem |
Hint |
Answers |
| 21.1 |
The first step is
to do the geometry. The angle between BD and vertical turns out to be 16.779 (be
sure you calculate this value and don't just use this hint). |
aD =
296 m/s2 (up) |
| 21.2 |
|
aBD =
9.99 rad/s2 (CCW)
aEx = 3.75 m/s2 (left)
aEy = 1.05 m/s2 (up) |
| 22.1 |
Only
the magnitudes are given to the right. |
a) |
| 22.2 |
You
will need to use kinematics to relate the acceleration of the string to
the acceleration of the center of gravity. |
a)
T = 0.197 lbf
b) alpha = 65.0 rad/s2 |
| 23.1 |
It will stop start to rotate
without slipping when vG = wr |
a) 3.22
ft/s2
b) 24.15 rad/s2
c) 1.60 s
d) vG = 9.86 ft/s
e) xG = 19.9 ft
f) |
| 23.2 |
|
a) a =
2.81 rad/s2 (CCW)
aG = 0.337 m/s2 (left)
b) ms =
0.22 |
HOMEWORK SET #9
| Problem |
Hint |
Answers |
| 24.1 |
|
aA = 2P/7m
aB = 22P/7m |
| 24.2 |
|
b) q =
54.7 |
| 25.1 |
Determine a as a
function of q and integrate for q = 0 to 90 degrees. |
w = 2.22 rad/s |
| 25.2 |
The easiest way to write the
position vector for bar BD is in terms of an angle with respect to the horizontal, b, where
b=sin-1(bsinq/L).
Be sure you know where this equation comes from. |
Answer for when
q = 180:
By = 0, Bx = 805 N (left)
Dy = 0, Dx = 426 N (right). A plot of Bx
was provided with the problem statement. |
HOMEWORK SET #10
| Problem |
Hint |
Answers |
| 28.1 |
The direction
of vrel will be along bar AD. |
a) 5.16 rad/s (CW)
b) 1.34 m/s 60
up from horizontal left |
| 28.2 |
|
a) 24.2 in/s 82.9up
from horizontal left
b) 64.4 in/s2 26.6down
from horizontal right |
| 29.1 |
You need to
solve the velocity problem first. You should find vrel =
200 mm/s (down). Don't forget the normal component of arel. |
20 rad/s2 CW |
| 29.2 |
Use the answer
you obtained for 28.1 for wBP
and vrel. For part c) include the slider in your system,
so the force at P will act perpendicular to bar AD. |
a) 64.2 rad/s2 CW
b) 24.4 m/s2 60
up from horizontal left
c) By = 57.6 N (up)
Bx = 27.5 N (left) |
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