MAE2323 Dynamics Homework 9 solutions Spring 2026 University of Texas Arlington

MAE2323 Dynamics Homework #9 (235pts) Due April 8, 2026


1 Homework #9 Solutions

Grade.

1.1 S5.25 (5pts) What is the system level sum of forces? The summation of all forces acting on a
system of rigid bodies.

1.2 S5.26 (5pts) Under what conditions is linear momentum conserved? When the sum of forces
on a system is equal to zero.

1.3 S5.27 (5pts) When is it useful to sum moments about a point other than the mass center?
If the arbitrary point has zero velocity. If this is true, we may also be able to eliminate some moments from the
equations of motion.

1.4 S5.28 (5pts) What is a key advantage of summing moments about the mass center? Only
rotational quantities appear in Euler’s Second Law, which does not occur when moments are summed about an
arbitrary point.

1.5 S5.30 (5pts) Under what circumstances is angular momentum conserved? When the sum of
moments about some point is equal to zero.

1.6 P5.23 (117pts) In Figure 5.70, a man will attempt to push a cabinet on a horizontal ice-
covered surface; you can assume that the surface is frictionless. The man weighs 180lbs and the
cabinet weighs 300lbs. At the instant shown in Figure 5.70, both the man and the cabinet are
at rest. Then the man will push on the cabinet. The open circles represent the mass centers
of each body. Over this short time, you can consider the man as a single rigid body. Find
the velocity of the man’s mass center if the cabinet begins to move toward the right at 3ft/s.
(answers: V = 5f t/s.)
Position Orientation: (37pts)


^
YN ^
YA
^
YB



hA
^ ^
A XA XB
N
x B
z

gmB
gmA
hB
^
XN
r
n p


Figure 1.0: Figure 5.70: Pushing Cabinet, Frames (6pts), points (3pts), coordinates (2pts), parameters (0pts)

Simplications: We will use conservation of linear momentum to solve this problem.
Rigid Bodies: This system involves two rigid bodies, the man and the cabinet. (1pts)
Inertial Reference Frame and Point: The inertial reference point is labeled N and the inertial
reference frame is N = (XN , YN ). (1pts)
Other Points and Frames: The body-attached points are A and B, which represent the mass
centers of each body. The body body-attached frames are A = (XA , YA ) and B = (XB , YB ).
(1pts)
Location Descriptions: (18pts)
LA = PNA , NA R  + geom. LB = PNB , NB R  + geom. (4pts)
(2.26) (2.26)


1

, MAE2323 Dynamics Homework #9 (235pts) Due April 8, 2026


(1.1) PNA = x XN PNB = z XN (4pts)
(2.9) (2.9)

^ ^
YN YA

 =


XA XN

N
=  YA = YN
(2.19a) 
AR



 ZA = ZN
XA ^
^ XN (5pts)

^ ^
YN YB

 =


XB XN

N
=  YB = YN
(2.19a) 
AR



 ZB = ZN
XB ^
^ XN (5pts)
Coordinates: The location descriptions reveal two coordinates, x and z . (1pts)
Constraints: All three coordinates can be set freely so no constraints are required. (1pts)
Degrees of Freedom (DOFs): 2 coordinates − 0 constraints = 2 DOFs (1pts)
(2.83)
Givens: mA = 180lbs, mB = 300lbs, N VNB (t2 ) = 3f t/s XN . (1pts)
Objectives: Find N VNA (t2 ). (1pts)
Velocity: (8pts)
We will need the velocity of point A,
N
VNA = = ẋ XN
N dPNA
(1.2) (4pts)
(3.8) dt (1.1)
and of point B,
N
VNB = = ż XN
N dPNB
(1.3) (4pts)
(3.8) dt (1.1)
Mass Properties: The total mass of the man is mA and that of the cabinet is mB . (4pts)
Forces and Moments: (16pts) + FBD (14pts)
In order to apply the conservation of linear momentum for a system boundary that only
includes both bodies, we need to consider the external forces acting on the system:
(∑F)A = (n + r − gmA ) YN (6pts)
(5.7)
(∑F)B = (p − gmB ) YN (4pts)
(5.7)
Therefore, the sum of forces is:
(1.4) (∑F) = (∑F)A + (∑F)B = (n + r + p − gmB − gmA ) YN (6pts)
(5.38)
We can see that there are no external forces in the XN direction, so we can use conservation
of linear momentum in that direction.
Solution: (38pts)
From the given information we obtain:
(1.5) N
VNB (t2 ) = ż(t2 ) XN = 3f t/s XN Ð→ ż(t2 ) = 3f t/s (2pts)
(1.3)
Since we are using the conservation of linear momentum, we need the linear momentum of
each body:
KA = mAN VNA = mA ẋ XN (4pts)
(4.4) (1.2)
KB = mB VNB = mB ż XN
N
(4pts)
(4.4) (1.3)
Therefore,
(1.6) K = KA + KB = (mA ẋ + mB ż) XN (6pts)
(4.14)
The statement of conservation of linear momentum
0 = ∫ 0 dt = ∫ (∑F) ⋅ XN dt = (K(t2 ) − K(t1 )) ⋅ XN
t2 t2
(8pts)
t1 (1.4) t1 (5.42)
Therefore:
0 = (K(t2 ) − K(t1 )) ⋅ XN = mA ẋ(t2 ) + mB ż(t2 ) − mA ẋ(t1 ) − mB ż(t1 ) (4pts)
(1.6)


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