ENGI 1313 Mechanics I - Memorial University of Newfoundland

Transcript ENGI 1313 Mechanics I - Memorial University of Newfoundland

ENGI 1313 Mechanics I
Lecture 38:
Examples of Friction Analysis
Shawn Kenny, Ph.D., P.Eng.
Assistant Professor
Faculty of Engineering and Applied Science
Memorial University of Newfoundland
[email protected]
Lecture 38 Objective

2
to illustrate the equilibrium analysis of rigid
bodies subjected to dry friction force by
example
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38
Miscellaneous Notes

Lecture 37 Notes Have Been Reposted


3
Minor corrections
Quiz and Mid-Term Marks Posted
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38
Example 38-01

4
The 800-lb concrete
pipe is being lowered
from the truck bed when
it is in the position
shown. If the coefficient
of static friction at the
points of support A and
B is μs = 0.4, determine
where it begins to slip
first: at A or B, or both at
A and B.
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38
Example 38-01 (cont.)

Three Analysis Cases

Impending motion at A
FA  sNA & FB  sNB

Impending motion at B
FA  sNA & FB  sNB

Impending motion at A & B
FA  sNA & FB  sNB
5
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38
Example 38-01 (cont.)

Draw FBD
FA
FB
NA
6
© 2007 S. Kenny, Ph.D., P.Eng.
W = 800lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-01 (cont.)

Assume Impending
Motion at A & B
FA  sNA & FB  sNB

Analysis
M
O
0
 15FA  15FB  0
FA  FB
 NA  NB
7
© 2007 S. Kenny, Ph.D., P.Eng.
FA
FB
NA
W = 800lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-01 (cont.)

Analysis
 Fx  0
NA  FB  W sin30   0
NA  sNA  W sin30  0
NA  NB  285.7 lb  286lb

F
y
0
FA  W cos 30   NB  0
FA
FA  800lbcos 30  285.7 lb  0
FA  407.1lb  407 lb
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© 2007 S. Kenny, Ph.D., P.Eng.
FB
NA
W = 800lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-01 (cont.)

Assumption
FA  sNA & FB  sNB

What is Calculated
FA  FB  407.1lb  407 lb
NA  NB  285.7 lb  286lb

Check Assumption
FA  407 lb  sNA  0.4286lb  114.4lb

9
 not correct
© 2007 S. Kenny, Ph.D., P.Eng.
FA
FB
NA
W = 800lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-01 (cont.)

Assume Impending
Motion at A
FA  sNA & FB  sNB
FA
FB
NA
10
© 2007 S. Kenny, Ph.D., P.Eng.
W = 800lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-01 (cont.)

Analysis
M
O
0
 15FA  15FB  0
FA  FB  s NA

F
x
0
NA  FB  W sin30   0
NA  sNA  W sin30  0
FA
FB
NA  285.7 lb  286lb
 FA  FB  sNA  114.3lb  114lb
11
© 2007 S. Kenny, Ph.D., P.Eng.
NA
W = 800lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-01 (cont.)

Analysis
NA  285.7 lb  286lb
FA  FB  114.3lb  114lb

F
y
0
FA  W cos 30   NB  0
114.3  800lbcos30  NB  0
NB  578.5 lb  579lb
FA
FB
NA
12
© 2007 S. Kenny, Ph.D., P.Eng.
W = 800lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-01 (cont.)

Assumption
FA  sNA & FB  sNB

What is Calculated
FA  FB  114.3lb  114lb
NB  578.5 lb  579lb

Check Assumption
FB  114lb  sNB  0.4579lb  231.6lb

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 Slip impending at A
© 2007 S. Kenny, Ph.D., P.Eng.
FA
FB
NA
W = 800lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-02

14
The man having a weight of
150 lb pushes horizontally
on the bottom of crate A,
which is stacked on top of
crate B. Each crate has a
weight of 100 lb. If the
coefficient of static friction
between each crate is μs1 =
0.8 and between the bottom
crate, his shoes, and the
floor is μs2 = 0.3, determine
if he can cause impending
motion.
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38
Example 38-02

Impending Motion?
A/B Interface
 B/Floor Interface
 Man/Floor Interface
 Potential impending
motion at all three
interfaces (rare)
 Crates tip

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© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38
Example 38-02 (cont.)

Assume Impending
Motion for Crate A
FBD Crate A
 Analysis

F

0
y
FA
NA  W  100lb


F
x
W = 100lb
NA
0
P A  s1NA  0.8100lb  80lb
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© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38
Example 38-02 (cont.)

Assume Impending
Motion for Crate B
FBD Crate B
 Analysis

F

0
y
FB
NB  W  200lb


F
x
W = 200lb
NB
0
P B  s2NB  0.3200lb  60lb
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© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38
Example 38-02 (cont.)

Assume Tipping
FBD Crate A & B
 Analysis

M  0
200lb1ft   P 3ft   0
P T

 66.7 lb
Minimum P
P  min80, 60, 66.7 lb  60lb
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© 2007 S. Kenny, Ph.D., P.Eng.
FB
W = 200lb
ENGI 1313 Statics I – Lecture 38
NB
Example 38-02 (cont.)

Force Exerted by
Man? P  60lb
W
min
Pmin
FBD
 Analysis

F

0
y
NM  W  150lb


F
x
FM
0
FM  Pmin  60lb
NM
FM  60lb  SNM  0.3150lb  45lb
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© 2007 S. Kenny, Ph.D., P.Eng.
 Impending motion
ENGI 1313 Statics I – Lecture 38
References
Hibbeler (2007)
 http://wps.prenhall.com/esm_hibbeler_eng
mech_1

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© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 38