TUTORIAL #3, 26 Sep. 2008
GNG 1105A DGD
Moment
Resultant Forces (2D)
R x = ∑ Fx
R y = ∑ Fy
F = Fx i + Fy j
F = Fx2 + Fy2
Fx = F cos θ x ⇒ cos θ x =
Fx
F
Fy = FSinθ x ⇒ Sinθ x =
Fy
1
F
Space: (2D and 3D) Forces, Resultant, and Equilibrium
y
,
θy
F
A
θx
θz
B
Fy
x
Fx
Fz
z
Fx = F cos θ x ⇒ cos θ x =
Fx
F
Fy = F cos θ y ⇒ cos θ y =
Fy
Fz = F cos θ z ⇒ cos θ z =
→
AB : cos θ x =
X B− X A
AB
F
cos θ y =
YB− YA
AB
Fz
F
cos θ Z =
Z B− Z A
AB
F = Fx2 + Fy2 + Fz2
AB = (X B − X A ) 2 + (YB − YA ) 2 + ( Z B − Z A ) 2
1 = cos 2 θ x + cos 2 θ y + cos 2 θ z
→
AB
= FAB .λ AB
F = Fx i + Fy j + Fz k => F = FAB .
AB
→
AB : cos θ x =
X B− X A
YB −Y A
ZB− Z A
, cos θ y =
, cos θ Z =
,
AB
AB
AB
AB = ( X B − X A ) 2 + (YB − YA ) 2 + ( Z B − Z A ) 2
2
Resultant Forces
R x = ∑ Fx
R = R x2 + R y2 + R z2
R y = ∑ Fy
R = R xi + R y j + R zk =
Rz =
∑F
R = ∑ Fx .i + ∑ F y . j + ∑ F z .k
z
R = R x2 + R y2 + R z2
y
θy
F
θx
θz
Fy
x
Fx
Fz
z
R = R xi + R y j + R zk =
R = ∑ Fx .i + ∑ F y . j + ∑ F z .k
Equilibrium
∑F
x
=0,
∑F
y
=0,
∑F
z
=0
3
Vector Product of two vectors:
V = P×Q
Q
θ
P
V = PQsin θ
V = P×Q
P×Q = - Q× P
P×(Q1+ Q2) = P×Q1+ P×Q2
j
k
i
i×i = 0
i×j = k
i×k = -j
j×i =
j×j =
j×k =
k×i =
k×j =
k×k =
V = P×Q
i
j
k
V = Px
Py
Pz
Qx
Qy
Qz
V = P×Q =i (Py . Qz - Pz . Qy) –j (Px . Qz - Pz . Qx) +k(Px . Qy - Py. Qx)
4
Mo
F
θ
d
r
Moment of a force about a point
Mo = r×F
rFsin θ = Fd
rFsin θ = Fd
V = P×Q
P×Q = - Q× P
r×(F1+ F2+ …) = r×F1+ r×F2+….
r = xi+yj+x k
F = Fx i+Fy j+Fz k
Mo = Mx i+My j+Mz k
Mx = y Fz –z Fy
My = z Fx –x Fz
Mz = x Fy –y Fx
5
6
7
∑M
∑M
∑F
D
x
∑F
y
c
= 1.6m.P sin 60 − 1.1mP cos 60 = 585Nm
= −0.6m.P cos 60 = −2.4m.B → B = 87.5N
→ P cos 60 = D cos θ → D =
350
cos θ
→ P sin 60 = B + D sin θ = B + 350 tan θ → D = 626N
Second Method Next Page:
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