Download MCE 263 Dynamics

MCE 263 Dynamics
Final Exam
PROBLEM # 1
Thursday, May 3, 2007
NAME
SECT.
INSTRUCTOR
Do work for this problem on this sheet or on continuation sheets. Do not use back of
sheets.
1. A ball is thrown upward from the 12-meter level in an elevator shaft with an initial
velocity of 18m/s. At the same instant, an open-platform elevator passes the 5-meter
level, moving upward with a constant velocity of 2m/s. Determine the following:
a. (10 points) Equation for the position of the ball as a function of time-show
your reference
b. (10 points) Equation for the position of the open platform elevator as a
function of time-show your reference
c. (20 points) When and where the ball passes the open platform elevator
d. (10 points) The relative velocity of the ball with respect to the open platform
elevator when the ball passes the platform
MCE 263 Dynamics
Final Exam
PROBLEM # 2
NAME
SECT.
Thursday, May 3, 2007
INSTRUCTOR
Do work for this problem on this sheet or on continuation sheets. Do not use back of sheets.
2. (50pts) A 75kg man stands on a scale in an elevator. The elevator is initially at
rest, then accelerates upward for a period of 3 seconds, due to the tension in the
hoisting cable T = 8300N. The total mass of the elevator, the man, and the scale is
750 kg.
a) (10pts) Draw the free-body-diagram of the entire system comprising the
elevator, with the man, and the scale inside it.
b) (10pts) Write the equation of motion for the elevator.
c) (5pts) Solve the equation of motion to find the vertical acceleration of the
elevator.
d) (10pts) Draw the free body diagram for the man (alone).
e) (10pts) Write the equation of motion for the man (alone).
f) (5pts) Solve for the reading R of the scale (in other words, the force the man exerts on the scale).
Solution
a)
FBD for system: T : hoisting force, up = 8300 N
W : weight of system, down = 750(9.81) N
b)
ΣFy = m ay
8300 – 7360 = 750ay
c)
ay = 1.257 m/s2
d)
FBD for man:
R : reaction force, up
W : weight of man, down = 75(9.81) N
e)
ΣFy = m ay
R – 736 = 75ay (ay is known: 1.257 m/s2 )
f)
R = 830 N
MCE 263 Dynamics
Final Exam
PROBLEM # 3
NAME
SECT.
Thursday, May 3, 2007
INSTRUCTOR
Do work for this problem on this sheet or on continuation sheets. Do not use back of sheets.
3. (50pts) A 2kg ball is thrown at a horizontal speed of 10m/s and strikes
a 6kg block traveling down the incline at 1m/s. The ball/block
coefficient of restitution is e = 0.6. The kinetic coefficient of friction is
µk = 0.4.
y
x
a) (10pts) Write the conservation of momentum equation for the ballblock system.
b) (5pts) Write the coefficient of restitution equation for the ball-block system.
c) (5pts) Solve for the velocities vAx and vBx of the ball and the block parallel to the incline right after
impact.
d) (5pts) Solve for the velocities vAy and vBy of the ball and the block perpendicular to the incline right
after impact.
e) (5pts) Draw the free body diagram for the block after impact.
f) (5pts) Write the equation of motion for the block.
g) (5pts) Find the friction force between the block and the plane.
h) (10pts) Using the principles of Work-and-Energy, find the distance that B slides up the plane before it
stops.
Solution
a)
Σm1 v1 = Σm2 v2
2(10 cos20o) – 6(1) = 2(vAx)2 + 6(vBx)2
(vAx)2 + 3(vBx)2 = 6.3969
b)
e = [(vBx)2 - (vAx)2] / [(vAx)1 - (vBx)1]
e = [(vBx)2 - (vAx)2] / [10cos20o – (-1)]
[(vBx)2 - (vAx)2] = 6.23816
c)
(vAx)2 = -3.0794 m/s
(vBx)2 = 3.1588 m/s
d)
(vAy)2 = (vAy)1 = -10sin20o = -3.4202 m/s
(vBy)2 = 0 m/s
e)
FBD for block:
W : weight, vertical down = 6(9.81) N
N : up, perpendicular to plane
F : friction, downward, parallel to plane
f)
ΣFy = m ay = 0
-6(9.81)cos20o + N = 0
g)
N = 55.31N
F = 0.4N = 22.1N
h)
T1 + Vg1 + ΣU12 = T2 + Vg2
½ mv21 + mgh1 + Fd = ½ mv22 + mgh2
½ (6)(3.1588)2 + 0 – 22.1d = 0 + 6(9.81)dsin20o
 d = 0.708m
MCE 263 Dynamics
May 3, 2007
Final Exam
Thursday,
PROBLEM # 2
NAME
SECT.
INSTRUCTOR
Do work for this problem on this sheet or on continuation sheets. Do not
use back of sheets.
5. The slender bar AB weighs 60 lbs and moves in the vertical plane, with its ends
constrained to follow the smooth (no friction) horizontal and vertical guides. The
30 lb force is applied at point A when the bar is an angle θ = 30 degrees. The bar
is initially at rest. Determine the following:
a. (5 points) Free Body Diagram
b. (10 points) Newton’s Equations of motion for bar AB based on the center
of gravity of the bar
c. (20 points) The x and y components of the acceleration of the center of
gravity of the bar AB in terms of the angular acceleration of the
bar AB
d. (15 points) Reaction forces at points A and B from the walls on the bar and the
angular acceleration of the bar AB