MAE2323 Dynamics
(282pts)
Homework #8
Due October 31, 2024
MAE 2323 DYNAMICS Homework #8 complete solution (282pts) University of Texas, Arlington Grade P5.12 and P5.16. S5.1 (5pts) State in words Euler’s First Law. The sum of external forces on a rigid body is equal to the change in linear momentum of the rigid body’s mass center. S5.2 (5pts) State in words Newton’s First Law. When observed from an inertial reference frame and point, a body at rest remains at rest, and a body in motion remains in motion, with a constant speed and direction, unless acted upon by an external force. S5.3 (5pts) State in words Newton’s Third Law. For every action there is an equal and opposite reaction. S5.4 (5pts) What is a reaction force? A force between two bodies that maintains a constraint. S5.5 (5pts) What is an internal force? A force between two bodies within a system. S5.9 (5pts) When are two sets of forces and moments considered to be equivalent? When they produce the same resultant force, as well as the same resultant moment about the same point. P5.7 (171pts) In Figure 5.54, two blocks, connected by ropes and springs, hang over a pulley. The smaller block has a mass of 0.005slugs, the particle C has a mass of 0.001slugs, the pulley has a mass of 0.01slugs, and the larger block has a mass of 0.4slugs. The solid lines passing over the pulley represent the rope. One side of the rope connects to a particle C, which is then connected to a spring. The pulley system was left out in the rain so that the axle passing through the center of the pulley has rusted, creating enough of a friction moment to stop the entire system from moving. Assume that there exists enough friction between the rope and pulley so that the rope does not slip over the pulley. Model the pulley as a particle located at its center. Find the value of the friction moment on the pulley. (answer: the magnitude of the friction moment is 12.69slug·ft /s .) 2
2
Position Orientation: (57pts) ^
^
YN YAYB D N y
x 2h
C
B
2w 1.6ft
k
A
1ft
XN X AXB
gravity
xo
r
2.1ft
z
Figure 5.54: Frames (6pts), points (5pts), coordinates (3pts), parameters (2pts)
Simplifications: The rod has negligable width. Rigid Bodies There are two rigid bodies in this problem, the rod and disk.
(1pts)
Inertial Reference Frame and Point: The inertial reference point is labeled N and the inertial ^ N, Y^N). reference frame is N = (X (1pts) Other Points and Frames: The body-attached points are A and B, which represent the mass centers ^B, Y^B). of the bodies. The body-attached frames are A =^(XA^ , YA) and B = (X (1pts) Location Descriptions: (24pts) } } LA = PNA, NR + geom. LB = PNB, NR + geom. (4pts) A
(2.26)
LC
= (2.32)
B
(2.26)
{PNC}
LD
= {PND} (2.32)
(2pts)