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1. A force F of magnitude 50 N is exerted onthe automobile parking-brake lever at theposition x = 250 mm. Replace the force by anequivalent force–couple system at the pivotpoint O.
EQUIVALENT FORCE–COUPLE SYSTEM
10sin25.025sin1.020sin5010cos25.025cos1.020cos50 xy FF
oM
+
jijiF
98.461.1720cos5020sin50
mNM o 29.17
20o
F
Mo=C=17.29 N.mResultants 1
2. Replace the force system acting on the post by a resultant force, and specify where its
line of action intersects the post AB measured from point B.
Resultants 2
3. The device shown is part of an automobile seat-back-release mechanism. The part is subjected
to the 4 N force exerted at A and a 300 Nmm restoring moment exerted by a hidden torsional
spring. Determine the y-intercept of the line of action of the single equivalent force.
jiR
jiFR
035.186.3
15sin415cos4
Total Moment about O (resultant couple):
Resultant Force:
mmNM
M
O
O
8.155
3004015cos41015sin4+
y
xO
d jiR
035.186.3
xR
yR
Ox MdR
mmy
mmd
d
3.40
3.40
8.15586.3
3
4. A machine component is subjected to the forces and couples shown. The component is
to be held in place by a single rivet that can resist a force but not a couple. Determine the
location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.
120 N
80 N
60 N
40 N.m
42 N.m
200 N
4
Determine the location of the rivet hole if it is to be
located (a) on line FG, (b) on line GH.
ij
ji
ji
FR
6080
70sin12070cos120
15sin20015cos200
jiFR
53.24414.212 Total Moment about G (resultant couple):
Resultant Force:
47.015cos20005.015sin200
19.070sin12047.070cos12013.06059.0804042
GM
mNMG 34.63
Resultants 5
Determine the location of the rivet hole if it is to be
located (a) on line FG, (b) on line GH.
Equivalent Force-Couple System at point G:
jiR
53.24414.212
mNMG 34.63
GGM
jiR
53.24414.212
(a) (b)
G
y
Equivalent Force-Couple System at G:
my 298.014.212
34.63
G
jiR
53.24414.212
jiR
53.24414.212
x
mx 259.053.244
34.63
F
H
Resultants 6
5. Replace the parallel force system acting on the plate by a resultant force and specify its
location on the x–z plane.
jjjjFR
10352
SOLUTION:
Resultant force
jik
jikjikM
CFrM
O
O
35.15.0
25.15.255.05.1
Sum of the moments about point O
kiM
kikikiM
O
O
1014
5.45.1355.25.7
O
kiizkx
kijkzix
RrMO
10141010
101410
jR
10
r
mzmx 4.1,1
Location on the x-z plane which the resultant force acts.
Resultants 7
6. Represent the resultant of the force system acting on the pipe
assembly by a single force and a couple at A .
kjiFR
100180120
SOLUTION:
j
A
A
jk
kiikM
CFrM
40
25.016050
1003.012025.0
kjM A
50100
Equivalent force-couple system at A
A
y
x
z
kjiR
100180120
kjM A
50100
Sum of the moments about point A
Resultant force
Resultants 8
7. The pulley wheels are subjected to the loads shown.
Determine the equivalent force-couple system at point O.
kjFR
5001000
jkijki
kjikjiM
CFrM
o
o
60018.02.140018.02.1
30024.06.020024.06.0
kjiM o
120030012
Equivalent force-couple system at point O
O
x
y
z
kjR
5001000
kjiM o
120030012
SOLUTION:
Sum of the moments about point O
Resultant force
Resultants 9
8. Two forces and a couple (applied to portion AB) are applied to the car’ s exhaust system as
shown. Replace the given force system with an equivalent force-couple system at D.
25 N.m
115 N
100 N
kjR
kjjFR
FF
5039.28
30sin10030cos100115
21
SOLUTION:
Resultant force
Resultants 10
25 N.m
115 N
100 N
CFrM D
kjiM
kjiM
D
D
868.179825.8
207.43568.4115243.94475.85
kjjkikikFr
505.86225.03.079.046.031.036.022
jikikFr
11569.046.031.036.011
ikFr
3.947.4311
kjiFr
568.4124745.8522
y
z
C
kjkj
C
2015375.0
3.0225.025
Equivalent force-couple system at D
kjiM
kjR
D
868.179825.8
5039.28
Sum of the moments about point D
B
A
11
9. Replace the three forces acting on the plate by a wrench. Specify the magnitude of theforce and couple moment for the wrench and the point P(x, y) where its line of actionintersects the plate.
kjiFR
800300500
Resultant force
Equivalent force-couple system at A
jjikjM
CFrM
A
A
300468004
kiM A
18003200
Sum of the moments about point
xy
z
AM
R
A
kjiFR
800300500
Equivalent force-couple system at A
kiM A
18003200
Reduction to wrench:
Unit Vector of Resultant force: kjikji
nR
808.0303.0505.0
95.989
800300500
13
0304000018003200800300500 kikji
(Couple is not perpendicular to the force. Force-couple system can be reduced to a wrench – positive wrench.)
Parallel component of the couple to resultant force:
mNkjikiM R 4.3070808.0303.0505.018003200//
Parallel component of the couple to resultant force in vector form:
kjikjiM R
64.568833.93055.1550808.0303.0505.04.3070//
(Wrench Resultants)
kjiM R
64.568833.93055.1550//
xy
z
RM
R
A
//RM
kjiR
800300500
Positive Wrench
Resultants 14
Normal component of the couple to resultant force in vector form
kjikiMMM RAR
64.568833.93055.155018003200//
kjiMMM RAR
64.388833.93045.1649//
Coordinates of the point in the xy plane through which the line of action of the wrench passes
xy
z
RM
R
A
//RM Positive Wrench
kjikjijyixMRr R
64.388833.93045.1649800300500
kjiiykyjxkx
64.388833.93045.1649800500800300
mxxj
myyi
16.133.930800
06.245.1649800
x
y
z
R
A
//RM
(Positive Wrench)
(1.16,2.06,0)
10. a) Reduce the force system to a force-couple system at O. b)Replace the force couple
system obtained by a wrench and determine the coordinates of the point in the yz plane
through which the line of action of the wrench passes.
Resultants 15
11. The force-couple system acting at O is equivalent to the wrench acting at A.
If and , determine . NkjiR
7001400600 mNM //R 1200
oM
kjikji
R
RnR
418.0835.0358.07001400600
7001400600
222
(Unit vector of resultant force)
kjikjinMM RRR
6.50110026.429418.0835.0358.01200////
kjikjijiRrM R
300021001400700140060023
kjiMMM RRO
6.350110986.1829//
SOLUTION:
Resultants 16
jiR
jiijiFR
jijiF
iF
jijiF
35201040
800150050027202040
800150017
81700
17
151700
500
272020405
43400
5
33400
3
2
1
kjiCFrM
kC
kjijikjFr
kijFr
kjiFr
jikjijijikFr
CFrM
A
A
6.11391770960
800
75075040080015005.05.0
1505003.0
6.48910201360
272020405.016.03.02720204017
8
17
1534.05.0
33
22
11
11
17
12. The pulleys and the gear are
subjected to the loads shown. For these
forces, determine the equivalent force-
couple system at point A.