Module 14: Moment of Iner=a
November 19, 2010
Module Content:
1. Center of gravity, center of mass, and centroid are similar concepts which describe specic
loca=ons in a body. The share similar approaches to their calc
Module 13: Centroid and Center of Gravity
November 15, 2010
Module Content:
1. Center of gravity, center of mass, and centroid are similar concepts which describe specic
loca=ons in a body. The share similar approaches
Module 12: Fric=on Fundamentals
November 1, 2010
Module Content:
1. Fric=on is the force of interac=on between two bodies in rela=ve (sliding) mo=on.
2. The impending mo=on condi=on and equa=ons are cri=cal to iden=fy for
Fundamental Problem 4.26
Replace the loading system by an
equivalent force and couple moment
ac8ng at point A.
ans.: FR = 108 N @ 68.2o, MR = 470 Nm (cw)
MAE 2300 Statics
E. J. Berger, 2010
P6- 1
Fundamental Problem 4.30
Re
Module 7: Equivalence & Distributed Loads
September 13, 2010
Module Content:
1. Equivalence calculaLons allow us to simplify a loading scenario on a structure.
2. Equivalence condiLons are ensured only if we saLsfy both f
Fundamental Problem 4.6
Determine the magnitude and
direc7on of the moment of the force
around point O.
ans.: MO = 1.06 kNm
MAE 2300 Statics
E. J. Berger, 2010
P4- 1
Fundamental Problem 4.12
Determine the moment about point O
Module 6: Moments and Equivalence
September 10, 2010
Module Content:
1. Cross products are frequently used in sta=cs for calcula=ons involving moments.
2. The moment about a par=cular axis is calculated using the scalar t
Module 5: Par=cles in 3D
September 6, 2010
Module Content:
1. Par=cle equilibrium analysis rests upon free body diagrams.
2. 3D equilibrium scenarios are approaches in exactly the same way as 2D problems.
3. Cross produ
Problem 2.90
Determine the magnitude and direc7on of the
resultant force. Determine the coordinate
direc7on angles of the resultant force.
ans.: FR = 822 N
= 72.8, = 83.3, = 162.0 (degrees)
MAE 2300 Statics
E. J. Berg
Module 4: Yes, Even More Vectors
September 1, 2010
Module Content:
1. Dot and cross product opera=ons also have specic physical meanings in sta=cs.
2. Par=cle equilibrium analysis rests upon free body diagrams.
Module Read
Fundamental Problem 2.18
Determine the resultant force ac9ng
on the hook. Express your result as
Cartesian vector components.
MAE 2300 Statics
E. J. Berger, 2010
P3- 1
Fundamental Problem 2.21
Express the force as a Cartesian
Module 3: More Vectors
August 30, 2010
Module Content:
1. Cartesian vectors are especially useful forms of vectors.
2. Many physical quan=es relevant to structural analysis can be formulated as vectors.
3. Dot and cross
Fundamental Problem 2.3
Determine the magnitude and
direc/on of the resultant force.
MAE 2300 Statics
E. J. Berger, 2010
P2- 1
Problem 2.4
Determine the magnitude and
direc/on of the resultant force.
MAE 2300 Statics
E. J. Berger, 2
Module 1: Welcome
August 25, 2010
Module Content:
1. [email protected] is your rst true engineering course, and it is in the category of engineering
mechanics. You will have two other engineering mechanics courses, strength of ma
MAE 2320 Dynamics Spring 2011
Project Assignment
Due Date: Wednesday May 4, 2011 (noon)
Overview of Video SoluHons. You have all seen the video solu/ons on the publishers website
(h6p:/www.prenhall.com/hibbeler). These v
Fundamental Problem 2.3
Determine the magnitude and
direc/on of the resultant force.
MAE 2300 Statics
E. J. Berger, 2010
P2- 1
Problem 2.4
Determine the magnitude and
direc/on of the resultant force.
MAE 2300 Statics
E. J. Berger, 2
The coordinate system Oxy is defined above.
First, we find the x and y components of vector F2 and F1
0
F1,x 200 cos 45 141.42 lb
F1,y
200
0
sin 45
141.42 lb
0
F 2,x 150
cos 60 75 lb
F 2,y 150
sin 60 129.9 lb
0
Computing the resultant vectors x and y co
Julie Kokinos
+y
F2
+x
F1
1. The coordinate system is set up in the above diagram with the positive xaxis pointing to the left and the
positive yaxis pointing upward. The xaxis is defined as opposite of