rigid bodies: equivalent system of forces
DESCRIPTION
Rigid Bodies: Equivalent System of Forces. the effects of forces on a rigid body. to replace the system of forces with a simpler equivalent system. In this chapter we will study. F. R 2. R 1. W. External and Internal Forces. The External Forces - PowerPoint PPT PresentationTRANSCRIPT
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 1
Rigid Bodies: Equivalent System of ForcesEquivalent System of Forces
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 2
• the effects of forces on a rigid body.
• to replace the system of forces with a simpler equivalent system.
In this chapter we will study
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 3
External and Internal Forces • The External Forces
represent the action of other bodies on the rigid body under consideration. They control the external behavior of the body.
• The Internal Forces Are the forces which hold together the
particles forming the rigid body.
F
R2R1 W
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 4
Principle of TransmissibilityEquivalent Forces
The conditions of equilibrium or motion o a rigid body will remain unchanged if a force F acting at a given point of the rigid body is replaced by a force F’,
F
=F’A
B
same magnitude and direction
provided on the same line of action
acting at a different point
F=F’
F
R2R1 W=
F
R2R1 W
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 5
Limitation
= =
= =
P1=-P2
P1=-P2
A BP1 P2
(a)
A BP1
P’2
(b)
A B
(c)
A BP2 P1
(d)
A BP1P’2
(e)
A B
(f)
While the principle of transmissibility may be used freely to determine the conditions of motion or equilibrium of rigid bodies and to compute the external forces acting on these bodies, it should be avoided, or at least used with care, in determining internal forces and deformations.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 6
Vector Product of Vectors
P
Q
V = P x Q
Line of action of V
The magnitude of V
V = P Q sin The sense of V
VRight Hand Rule
Right hand’s fingers show the direction of P and when you curl your fingers toward Q the direction of your thumb is the sense of V.
VAlso referred as Cross Product of P and Q.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 7
Properties
P
QQ’
V
V = P x Q = P x Q’
The magnitude of V is the area of the parallelogram.
The vector products are not commutative Q x P P x Q
but Q x P = -(P x Q)
The distributive property holds P x (Q1 + Q2) = P x Q1 + P x Q2
The associative property does not apply P x (Q x S) (P x Q) x S
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 8
x
y
z
Vector Products Expressed in Terms of Rectangular Components
i
j
k
0
k
- j
- k
0
i
j
- i
i x i =
i x j =
i x k =
j x i =
j x j =
j x k =
k x i =
k x j =
k x k = 0
i
jk
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 9
V = P x Q
V = P x Q = (Px i + Py j + Pz k) x (Qx i + Qy j + Qz k)
V = (Py Qz - Pz Qy ) i + (Pz Qx - Px Qz) j + (Px Qy - Py Qx ) k
zyx
zyx
QQQ
PPP
kji
V
yx
yx
zyx
zyx
PP
QQQ
PPP
jikji
Repeat first and second columns to the right. The sum of the products obtained along the red line is then subtracted from the sum of the product obtained along the black lines
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 10
O
Moment of a Force about a Point
F
r
d A
Mo
Moment of F about O
Mo = r x F
Mo
Mo = r F sin = F d
The magnitude of moment of F about O.
The magnitude of Mo measures the tendency of the force F to make the rigid body rotate about a fixed axis directed along Mo.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 11
Problems Involving Only Two Dimensions
O
d
F
Mo
Counterclockwise
Mo = + F d
points out of screen
O
d
F
Mo
Clockwise Moment
Mo = - F d
points into the screen
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 12
Varignon’s TheoremThe moment about a given point O of the resultant of several concurrent forces is equal to the sum of the moments of the various forces about the same point O. (VarignonVarignon (1654-1722))
x
y
z
OF1
F2
F3
F4
F5
r
r x ( F1+F2+ … ) = r x F1 + r x F2 + …
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 13
Rectangular Components of the Moment of a Force
x
y
z
O
Fy j
Fx i
Fz k
y j
z kx i
r = x i + y j + z k
r
F = Fx i + Fy j + Fz k
Mo= r x F
Mo = Mx i + My j + Mz k
Mx = y Fz – z Fy My = z Fx – x Fz Mz= x Fy – y Fx
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 14
x
y
z
O
Fy j
Fz k
Fx i
(yA-yB) j
rA/B AB
(z A-z B
) k (xA-xB) i
MB= rA/B x F = (rA-rB) x F
zyx
BABABA
FFF
zyx ///
kji
MB
xA/B = xA - xByA/B = yA - yB
zA/B = zA - zB
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 15
2-D problems
x
y
z
O
F
Fx i
Fy j
A (x, y, 0)
r
x i
y j
Mo = Mz kx
y
z
O
F
Fx i
Fy j
A
r A/B
MB = MB k
B(xA-xB) i
(yA-y
B)
j
Mo = (x Fy - y Fx) k
Mo = Mz = x Fy - y Fx
MB = (xA-xB) Fy - (yA-yB) Fx
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 16
Sample Problems
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 17
A 300-N vertical force is applied to the end of a lever which is attached to a shaft at O. Determine (a) the moment of the 300-N force about O(b) the magnitude of the horizontal force applied at A which creates the same moment about O.(c) the smallest force applied at A which creates the same moment about O.(d) how far from he shaft a 750-N vertical force must act to create the same moment about O.(Beer and Johnston)
O
300 N
60o
500
mm
A
A
60o
500
mm
dO
MO
a) Moment about O
d = (0.5 m) cos 60o = 0.25 m
MO = F d =(300 N) (0.25 m) = 75.0 Nmor
r
r = (0.5 m cos 60o) i + (0.5 m sin 60o) j = (0.25 i + 0.433 j) m
300 N
F = - 300 j N
MO = r x F = (0.25 i +0.433 j) x (-300 N) j = -75.0 k Nm
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 18
A
60o
500
mm
d
O
MO
b) Horizontal Force
d = (0.5 m) sin 60o = 0.433 m
MO = F d =(F) (0.433 m) = 75.0 Nm
or
r
r = (0.5 m cos 60o) i + (0.5 m sin 60o) j = (0.25 i + 0.433 j) m
F
F = F i N
MO = r x F = (0.25 i +0.433 j) x (F ) i = -75.0 k Nm
F = 75.0 Nm / 0.433 m = 173.2 N
- 0.433 F k Nm = -75.0 k Nm
F = 173.2 N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 19
A
60o
500
mm
O
MO
c) Smallest Force
d = (0.5 m)
MO = F d
F
F = 75.0 Nm / 0.5 m = 150.0 N
Since Mo = F d, when d is max F is the smallest.
75.0 Nm = F (0.5 m)
F
30o
d) 750 N vertical force
75 Nm = (750 N) d
A
60o
500
mm
O
MO
750 N
d
B
OB cos 60o = d
d = 0.1 m
OB = 0.1 m / cos 60o = 200 mm
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 20
A force of 800 N acts on a bracket as shown. Determine the moment of the force about B. (Beer and Johnston)
60o
B
A
200 mm
160
mm
800 N
MB = rA/B x F
800 N
B
Fx = (400 N) i
Fx = (693 N) j
A
rA/B
rA/B = (- 0.2 i +0.16 j ) m
F = (800 N) cos 60o i + (800 N) sin 60o j
= 400 i + 693 j
MB = (-0.2 i + 0.16 j) x 400i + 693 j
64.0 Nm) k = (- 138.6 Nm) k -
= (- 202.6 Nm) k MB = 203 Nm
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 21
A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A. (Beer and Johnston)
MA = rC/A x F
rC/A = (0.06 i +0.075 j ) m F = -(200 N) cos 30o j + (200 N) cos 60o k
= -173.2 j + 100 k
MA = (0.06 i + 0.075 j) x (-173.2 j + 100 k) - 6 j = -10.39 k
MA = 14.15 Nm
MA = 7.5 i – 6 j –10.39 k
x
y
z
60 mm
25 mm
50 mmA
B C
30o
60o
200 N
A(0,-50,0) C(60,25,0)
+ 7.5 i
x
y
z
MAx
MAz
MAyMA
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 22
Scalar Product of Two Vectors
P
Q
P • Q = P Q cos
scalar product
or
dot product
The result is a scalar
PropertiesCommutative P • Q = Q • P
Distributive P • (Q1 + Q2) = P • Q1 + P • Q2
Associative P • (Q • S) = (P • Q) • S?
P • (Q • S) and (P • Q) • S does not have a meaningX
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 23
Moment of a Force about a Given Axis
x
y
zO
L
•
A
FMO
C
Let OL be an axis through O; we define the moment MOL of F about OL as the projection OC of the moment on the axis OL.
MOL = • MO = • (r x F)
the moment MOL of F about OL measures the tendency of the force F to impart to the rigid body a motion of rotation about a fixed axis OL.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 24
Sample Problem
A B
C
F
D
E
PG
a
A cube of side a is acted upon by a force P as shown. Determine the moment of P
a) About Ab) About the edge ABc) About the diagonal AG of the cube
A B
C
F
D
E
G x
y
z
O ij
k
a)Moment about A
rF/A = ai – aj
kjP22
PP
kjikjjiPrM F/AA 22
PaPa
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 25
b) Moment about AB
22
PaPaM AB kjiiMi A
A B
C
F
D
E
PG
c) Moment about diagonal AG.
kjikjiAG
λ
3
1
3a
aaa
AG
6
111623
1 PaPaPaM AG kjikjiMλ A
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 26
Two forces F and –F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple.
Moment of a Couple
F-F
B
A
O x
y
z
rA
rB
rA x F + rB x (- F) = (rA – rB ) F
rA – rB = r
r M = r x F
The vector is called the moment of the couple; it is a vector perpendicular to the plane containing the forces and its magnitude
M = r F sin d = r sin
d
M = F dSince the r is independent of the choice of origin O, taking moment about another point would not change the result. Thus, the M of a couple is a free vector which may be applied at any point.
M
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 27
Equivalent Couples
O x
y
z
10
0 m
m100 mm
M
150 mm
200 N
200 N100 mm
O x
y
z
M
300 N
300 N
O x
y
z
M
100 mm
300 N
300 NTwo couples having the same moment M are equivalent, whether they are contained in the same plane or in parallel planes
This property indicates that when a couple acts on a rigid body, it does not matter where the two forces forming the couple act, or what magnitude and direction they have. The only thing which counts is the moment of the couple. Couples with the same moment will have the same effect on the rigid body.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 28
P1
P2
Addition of Couples
P1
P2
Consider two intersecting planes P1 and P2.
F1
- F1
A
B
F2
- F2
r
R
-RM = r x R = r x (F1 + F2)
M = r x F1 + r x F2
M = M1 + M2
M1
M2
M
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 29
Couples May Be Represented by Vectors
O x
y
z
dF
-F
O x
y
z
=
M
M = F d
= O x
y
z
M
O x
y
z
= Mx
My
Mz
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 30
Resolution of a Given Force into a Force at a Different Point and a Couple
F
Or
A
=
F
-F
=
O
MO
F
F
O
r
A
A
A force-couple system
The force-couple system obtained by transferring a force F from a point A to a point O consists of F and a couple vector MO perpendicular to F.
Conversely, any force-couple system consisting of a force F and a couple MO which are mutually perpendicular may be replaced by a single equivalent force.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 31
Sample Problem (Beer and Johnston)
x
y
z
100 N
100 N
150 N
150 N
240 mm
140 mm18
0 m
m18
0 m
m
A
D
C
E
BDetermine the components of the single couple equivalent to the two couples shown
100 N100 N
Mx = - (150 N) (0.360 m) = - 54.0 N.m
My = + (100 N) (0.240 m) = 24.0 N.m
Mz = + (100 N) (0.180 m) = 18.00 N.m
M = - (54.0 N.m) i +(24.0 N.m) j + (18.0 N.m) k
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 32
Sample Problem (Beer and Johnston)
O
150 mm
300
mm
60 mm
400 N200 N
200 N
60o
B Replace the couple and force shown by an equivalent single force applied to the lever. Determine the distance from the shaft to the point of application of this equivalent force.
260
mm
O150 mm-(24 N . m) k
F = -(400 N) j
= O-(24 N . m) k
-(400 N) j
-(60 N . m) k
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 33
O
60o
O
-(400 N) j
-(84 N . m) k
-(400 N) j
-(84 N . m) k = OC x F
=
C
= ( OC cos 60o i + OC sin 60o j ) x ( -400 N) j
-(84 N . m) k =
cos 60o (400 N) OC =
84 N . m
- OC cos 60o (400 N) k
= 0.42 m = 420 mm
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 34
Reduction of a System of Forces to One Force and a Couple
F2
O r2
A2
A3
A1
F3
F1
r1
r3
=
F2
O
F3F1
M1
M2
M3
=
O
RMO
R
R = F MOR = MO = (r x F)
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 35
Equivalent System of ForcesTwo systems of forces F1, F2, F3 , etc., and F’1, F’2, F’3, etc., are equivalent if, and only if, the sums of the forces and the sums of the moments about a given point O of the forces of the two systems are, respectively, equal.
F = F’ and MO = M’O
Fx = F’x Fy = F’y Fz = F’z
Mx = M’x My = M’y Mz = M’z
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 36
Sample Problem (Beer and Johnston)
A 4.8 m beam is subjected to the forces shown. Reduce the given system of forces to
a) An equivalent force couple system at Ab) An equivalent force couple system at Bc) A single force or resultant
a) An equivalent force couple system at A
R = F = (150 N)j – (600 N)j + (100 N)j –(250 N)j = -(600 N)j
150 N
600 N 100 N250 N
1.6 m 1.2 m 2 m
A B
MRA = (r x F) = (1.6i) x (-600j) + (2.8i) x (100j) + (4.8i) x (-250j) = - (1880 N.m)k
A B
- (600 N)j
- (1880 N)k
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 37
b) An equivalent force couple system at B
MRB = MR
A + BA x R = - (1880 N.m)k + (- 4.8m)i x (-600 N)j
= - (1880 N.m)k + (2880 N.m)k = (1000 N.m)k
A B
- (600 N)j
- (1880 Nm)k (2880 Nm)k
A B
- (600 N)j
(1000 Nm)k
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 38
c) A single force or resultant
A B
- (600 N)jx
r x R = MRA
(x)i x (-600 N)j = - (1880 N.m)k
- x(600 N)k = - (1880 N.m)k
x = 3.13 m
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 39
Sample Problem (Beer and Johnston)
R = F MRO = (r x F)
A square foundation mat supports the four columns shown. Determine the magnitude and point of application of the resultant of the four loads.
r, m F, kN r x F, kN•my
x
z
O -200j
2.5 m
z
y
x
A
B
CO
2 m 3 m
2.5 m
200 kN 60 kN
40 kN100 kN
0 0
-60j5i -300k
-40j5i + 2.5k 100i – 200k-100j2i + 5k 500i – 200k
R = -400j MRO = 600i –700k
-(400 kN)j
(600 kNm)i-(700 kNm)k
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
19.04.23 Dr. Engin Aktaş 40
y
x
z
O
-(400 kN)j
xi
zk
r x R = MOR
(xi + zk) x (-400j) = 600i – 700k
- 400xk + 400zi = 600i – 700k
- 400x = – 700
x = 1.750 m
400z = 600
z = 1. 500 m