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| These answers are provided for
you to check your work. If you get one of these answers and it does not
follow from your work the graders have been told to give you a ZERO
for the problem. The answers provided are not guaranteed to be correct.
If you get a different answer be sure to talk to your professor and
email richard.onyancha@rose-hulman.edu
with any corrections. In the answers provided sometimes only the magnitude is given and not the direction. For your homework be sure to include the magnitude and direction when appropriate. |
| Problem Set P-01 | ||
| problem | hint | selected answers |
| 3.4 |
You will need to separate variables and integrate. |
depth
=
6.98
miles |
| 3.5 |
VB/A=109.5i+20.29j km/hr rB/A=2.96 km |
|
| Problem Set P-02 | ||
| problem | hint | selected answers |
| 3.11 |
Be careful
with the directions you use for dependent motion and the direction of
your acceleration on your kinetic diagram. |
T1 = 13.42 N T2 = 24.78 N SA = 6.20 m |
| 3.12 |
The direction of aB/A is along the incline. | vB/A = 3.74 m/s down at 20 degrees |
| 3.14 |
When the car loses contact the normal force goes to zero. | rho = 668 ft, N = 120 lbf |
| 3.16 |
Remember the normal force is zero when the block leaves the incline. | theta = 27.5º, x = 3.81 ft |
| 3.24 |
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| 3.26 |
No hint |
Vr= 4.94 ft/s Vθ=7.5 ft/s N=2.30 Ib |
| Problem Set P-03 | ||
| problem | hint | selected answers |
| 3.29 | Be sure to factor in the static deflection of the spring due to plate B's weight. | h = 0.072ft k = 72.2 Ib/ft |
| 3.30 |
Be careful with the signs of
your velocities. |
e = 0.923 h1 = 1.28 m |
| 3.33 |
Be sure to analyze what
happens normal and tangential to the plane of contact. |
θ = 62.7º W= -0.14Vo^2 J I =0.36Vo Ns |
| 3.35 |
Analyze the impact in the
horizontal direction. |
Vo = 13.8 m/s |
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| Problem Set P-04 | ||
| problem | hint | answer |
| 4.3 |
TBE = 14.4 Ib (Compression) TCF = 3.98 Ib (Compression) a = 20.7 ft/s2 |
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| 4.5 |
a) T = 4.90 lbf b) F = 1.37 lbf (to the left) |
|
| TOP | ||
| Problem Set P-05 | ||
| problem | hint | answer |
| 4.1 |
No hint |
θ1 = 165
rev θ2 = 2200 rev |
| 4.8 |
No hint. |
aB = 15.7
rad/s2 aC = 5.24 rad/s2 (on B) aE = 3.18 m/s2 at an angle of 14.3 deg. up from left horiztonal. (on C) aE = 1.29 m/s2 at an angle 37.4 deg. up from horizontal right. |
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| 4.9 |
To maximize omega when the object is in the vertical position, find omega2 as a function of Lcg; differentiate with respect to Lcg; set equal to zero and solve for Lcg.A copy of your lab 2 worksheet needs to be attached to the solution to this problem. | Lcg = 0.1057 m |
| 4.10 | The spring is stretched at both positions. | omega = 11.1 rad/s CCW |
| 4.11 |
Treat the system has having two centers of gravity | alpha = 43.6 rad/s2 (CW) Cx = 21 N (left) Cy = 54.6 N (up) |
| TOP | ||
| Problem Set P-07 | ||
| problem | hint | answer |
| 4.21 |
Using IC's is the easiest way to do this problem. | 0.48 m/s to
the right |
| 4.26 |
Final energy include KE
for all bodies (2 terms each for rods), and PE for rods. Initial
includes spring energy only. Final energy include KE for all bodies (2
terms each for rods), and PE for rods. Initial includes spring energy
only. |
|vC| = 1.703 m/s |
4.29 |
Ball rolls without slip |
h = 2/5r |
4.31 |
Don't forget the impact |
solution |
|
| Problem Set P-08 | ||
| problem | hint | answer |
| 4.32 | 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 |
| 4.33 |
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 omegacyl. 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 |
| 4.35 | The translational acceleration of the mass center is zero. | a) aB = 5407 ft/s2 (down) b) aC = 5407 ft/s2 (up) c) aD = 5407 ft/s2 (down at 60°) |
| 4.36 | Relate the
acceleration of the mass center to both ends of the rod |
|
| 4.37 | No hint | aD = 296 m/s2 (up) |
| 4.38 | No Hint |
alphaBD = 9.99 rad/s CCW aEx = 3.75 m/s2 (left) aEy = 1.05 m/s2 (up) |
| Problem Set P-09 | ||
| problem | hint | answer |
| 4.41 | Only the magnitudes are provided in
the answers shown on the right |
aA = 18.4 ft/s^2, aB
= 9.2 ft/s^2 |
| 4.42 | T = 0.197 Ibf alpha = 65 rad/s^2 CW |
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| 4.43 | It will start to rotate without slipping when vG = omega r | 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) answer not given |
| 4.44 | No hint |
a) alpha = 2.81 rad/s2 (CCW), aG = 0.337 m/s2
(left) b) µs = 0.22 |
| 4.47 | no hint | aA = 2P/7m aB = 22P/7m |
| Problem Set P-10 | ||
| problem | hint | answer |
| 4.55 | Attach the body (rotating) frame to arm AD at A. Attach the fixed frame to bar BP at B. Rotate the body frame velocity to line up with the fixed frame (horizontal-vertical). | omega_BP = -5.17 rad/s k vrel = 1.344 m/s |
| 4.56 | Use frames with origins at A, and both oriented horizontal-vertical. | vB = (-0.25i + 2j) ft/s aB = (4.8i -2.4j) ft/s^2 |
| 4.57 | Don't forget the relative acceleration will have a component to the left since it is traveling in a curved path. | 20 rad/s2 CW |
|
Prof. Richard Onyancha Last modified: Wed Jan 4 2012 |