MAE 2600 (SP2009)
Homework #7 (Section 12B-4)
Name:_
Due: Monday, 2/9/09
ID#: (
)
H7-1. At the instant shown, the bicyclist at A is traveling at 5 m/s around the curve on the race
track while increasing his speed at 0.3 m/s2 . The bicyclist at B is travel
MAE 2600 (FS2013)
Homework #1 (Section 12A-1)
Name:_
Due: Friday, 8/23/13
ID#: (
)
H1-1. The velocity of a particle on a straight line is given by v = (3t2 - 15t + 12) ft/s, where t is
in seconds. Draw a path diagram from t = 0 to t = 5 s and determine th
MAE 2600 (FS2013)
Homework #4 Solutions
H4-1.
Given: at = v = (- 9.81 sin ) m/s2
v dv = at ds
v dv = - 9.81 sin (r d) where r = 0.5 m
v
v dv = - (0.5)(9.81) sin d
0
2
(1/2)(v2 - 4) = 4.905 [cos ] 0
v2 = 9.81(cos - 1) + 4 = 9.81 cos - 5.81
v = (9.81 cos -
MAE 2600
Ch. 13 Kinetics of a Particle: Force and Acceleration
Ch. 13 Kinetics of a Particle: Force and Acceleration
Section 13-1: 13.1, 13.2, 13.4
Section 13-2: 13.5
Class Note for Section 13-3:
13.6 Equations of Motion: Cylindrical Coordinates
F = m a
F
MAE 2600 (FS2013)
Homework #3 Solutions
H3-1.
horizontal motion: (at the slope)
+ x = xo + (vo)x t x = 0 + (3/5)(8) t = 4.8 t
(1)
vertical motion: (at the slope)
+ y = yo + (vo)y t + (1/2)(- g) t2
y = 20 + (4/5)(8) t + (1/2)(- 9.81) t2 = 20 + 6.4 t - 4.90
MAE 2600 (FS2013)
Homework #2 (Section 12A-2)
Name:_
Due: Monday, 8/26/13
ID#: (
)
H2-1. A car starts from rest and is subjected to a constant acceleration of ac = 10 ft/s2 for
0 t < 6 s. The brake is then applied, which causes a deceleration at the rate
MAE 2600 (FS07)
TEST #4
Name:_
ID#: (
)
This is a closed book exam. Show your work clearly and concisely!
Indicate answers and units. You have a full 100 min. period for the exam.
1. The 2-kg gear is at rest on the surface of a gear rack. If the rack is s
MAE 2600 (FS07)
TEST #4 SOLUTIONS
1.
+ Fx = m(aG)x : FA = 2a G
(1)
CCW+ MG = IG : FA(0.4) = 2(0.3)2 (2)
(Note: is positive and a CCW rotation.)
(1)&(2) aG = 0.225 (3)
(Note: a G & are positive numbers and
magnitudes of acceleration and angular acceleratio
MAE 2600 (FS2013)
Homework #9 (Section 13-3)
Name:_
Due: Friday, 9/13/13
ID#: (
)
H9-1. Rod OA rotates counterclockwise with a constant angular velocity of = 4 rad/s.
The double collar B is pin-connected together such that one collar slides over the rotat
MAE 2600 (FS2013)
Homework #7 (Section 13-1)
Name:_
Due: Monday, 9/9/13
ID#: (
)
H7-1. When the 30-lb block A is released from rest, determine the acceleration of the 12-lb
block C and the tension (TAB & TBC) in each cable. The weight of block B is 6 lb a
MAE 2600 (FS2013)
Homework #4 (Section 12B-1)
Name:_
Due: Friday, 8/30/13
ID#: (
)
H4-1. The block B has a velocity of vA = 2 m/s when it reaches point A. If its speed is
decreased at a rate of v = (- 9.81 sin ) m/s2 along a circular path from A, determin
MAE 2600 (FS2013)
Homework #6 (Section 12B-3)
Name:_
Due: Friday, 9/6/13
ID#: (
)
H6-1. When the block A is released from rest, it moves the block B 20 ft up in t = 5 s with a
constant acceleration. Determine the acceleration of block A and B during this
MAE 2600 (FS2013)
Homework #5 (Section 12B-2)
Name:_
Due: Wednesday, 9/4/13
ID#: (
)
H5-1. Using a rotating arm, a particle is forced to move along a slotted path r = (0.4/) m,
where is in radians. If the angular position of the arm is = (0.8 t2) rad, whe
MAE 2600 (FS2013)
Homework #3 (Section 12A-3)
Name:_
Due: Wednesday, 8/28/13
ID#: (
)
H3-1. The ball is thrown from the tower with a velocity of 8 m/s as shown. Determine the x
and y coordinates to where the ball strikes the slope.
H3-2. The man stands 50
MAE 2600 (FS2013)
Homework #2 Solutions
H2-1.
The equation of the line for 6 t < t1: a = (- 1/2.5)t + C
a = 0 at t = 6 0 = (- 0.4)(6) + C C = 2.4 a = - 0.4t + 2.4
v2 - v1 = v =
t2
t1
a dt (the area under the a-t curve between t1 and t2)
v2 = v1 = 0, A1 =
MAE 2600 (FS2013)
Homework #1 Solutions
H1-1.
v = 3t2 - 15t + 12 = 3(t2 - 5t + 4) = 3(t - 1)(t - 4)
v = 0 at t = 1 s and t = 4 s
At t = 0, s = 0 (given)
v = ds/dt ds = v dt
s
t
ds = (3t2 - 15t + 12) dt
0
0
s - 0 = t3 - 7.5t2 + 12t
Path Diagram
s = t3 - 7
MAE 2600 (SP12)
TEST #1
Name:_
ID#: (
)
This is a closed book exam. Show your work clearly and concisely!
Indicate answers and units. You have a full 50 min. period for the exam.
1. Cars A and B are traveling around the circular race track. At the instant
MAE 2600 (FS12)
TEST #1 SOLUTIONS
1.
At B (y = 25 ft): x = 25(1.6) + 18 = 58 ft
(A B):
+ x = xo + (vo)x t 58 = 0 + (3/5)vA t
+ y = yo + (vo)y t + (1/2)(- g) t 2
25 = 80 + (4/5)vA t + (1/2)(- 32.2) t 2
55 + 0.8vA t - 16.1 t2 = 0 (2)
(1)
From Eqs. (1) & (2)
MAE 2600 (FS12)
TEST #1
Name:_
ID#: (
)
This is a closed book exam. Show your work clearly and concisely!
Indicate answers and units. You have a full 50 min. period for the exam.
1. The ball is thrown from the tower with an unknown speed v A and strikes t
MAE 2600 (FS2013)
Quiz #9 Solutions
Q9A. The forked rod is used to move the smooth 5-lb particle
around the horizontal path in the shape of a limacon,
r = (2 + cos ) ft. If at all times = 2 rad/s, determine the force
of the rod on the particle at the inst
MAE 2600 (FS2013)
Quiz #8 Solutions
Q8A. A toboggan and rider have a total weight of 100 lb
and travel down along the slope defined by the equation
y = 0.02 x2 for which the coefficient of kinetic friction is
k = 0.2. At the instant x = 40 ft, the tobogga
MAE 2600 (FS2013)
Quiz #6 Solutions
Q6A. Block A is moving downward at vA = 0.6 m/s at the instant shown with a constant
acceleration of 0.2 m/s2. When block A has moved 4 m down, determine the velocity of block B.
For block A only:
v2 = vo2 + 2 ac (s - s
MAE 2600 (FS2013)
Quiz #7 Solutions
Q7A. A Force of F = 20 N is applied to the cord at B. Determine the acceleration of the 8-kg
block A. Neglect the mass of the pulleys and cord.
For block A only:
+ Fy = m ay:
80 - 8(9.81) = 8aA aA = 0.19 m/s2 Ans.
Q7B.
MAE 2600 (FS2013)
Quiz #5 Solutions
Q5A. A particle is moving along a circular path having a 2-m radius. Its position as a function
of time is given by = (2 t2) rad, where t is in seconds. Determine the radial and transverse
components of the particles ve
MAE 2600 (SP2016)
Homework #9 Solutions
H9-1.
At = 125, = 2 rad/s, = 3 rad/s2
r = 2.8(2 - cos ) = 2.8(2 - cos 125) = 7.206
r = 2.8 sin = 2.8 sin 125 (2) = 4.587
r = 2.8(cos 2 + sin )
= 2.8[(cos 125)(2)2 + sin 125 (3)] = 0.4568
ar = r - r 2 = 0.4568 - 7.20
MAE 2600 (SP2016)
Homework #11 (Section 14-2)
Name:_
Due: Monday, 2/22/16
ID#: (
)
H11-1. If the coefficient of kinetic friction is k = 0.2 between the slope and block B, determine
the velocity of 18-kg block A and 6-kg block B after the block B moves 1.2
MAE 2600 (SP2016)
Homework #9 (Section 13-3)
Name:_
Due: Friday, 2/12/16
ID#: (
)
H9-1. Rod OA rotates counterclockwise with = 2 rad/s and = 3 rad/s2 at = 125. The
double collar B is pin-connected together such that one collar slides over the rotating rod