Lecture Engineering Mechanics: statics

1. DEVAPRAKASAM DEIVASAGAYAM
Professor of Mechanical Engineering
Room:11, LW, 2nd Floor
School of Mechanical and Building Sciences
Email: devaprakasam.d@vit.ac.in, dr.devaprakasam@gmail.com
ME202: Engineering Mechanics (3:1:0:4)
Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933

2. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Problems

3. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Solution
TABy
TABx
TAB
M
N
P
21
75 72
MC= TABx (y) +TABy (x)
MC= (0.028)TABx + (0.021) TABy
MC= (0.028)[ (7/25) x1500] + (0.021) [(24/25)x1500]
MC= 42.0 N.m (CCW )

4. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Problems
A B
C E D
0.6m
0.5
m
0.85 m
W
0.6m
0.25m

5. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Solution
A B
C E D
0.6m
0.5 m
0.85 m
W
0.6m
0.25m
W= 784.8N
mNMM EE .2.19625.08.784: 
    mEBd 986.085.05.0)(
22

NF
FmNMM
B
BEE
954.198
)986.0(.2.196:


F



5.5990 
 

6. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Problem

7. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Solution
200lb
a) MB= rC/B x 200 lb= 4in x 200 lb
MB= 800 lb . in
b) MB= rA/B x P= rA/B P sin θ
800 lb . In = 18 in x P sin 60°
lbP 3.51
60sin18
800


 
c) MB= rA/B x P= rA/B P sin θ
For minimum P, θ=90°
;4.44
90sin18
800
lbP 

 