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CH 2: STATICS OF PARTICLES
Force component
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Equilibrium of a particle
Chapter Objectives:
To show addition of forces (resultant force)
To resolve forces into their components
To express force and position in Cartesian vectors.
To analyse the equilibrium of forces acted on a particle
Introduction to Force Vectors
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Force
» the action of one body on another.
» a vector.
» characterised by points of application,
magnitude and direction.
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»represented by P, P or P
45º
Vector Addition
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Parallelogram Diagram
P+Q=Q+P
P+Q≠P+Q
P – Q = P + (-Q)
R
P
O
Q
R
c
A
B
b
a
C
Law of Sinus
a/sin A = b/sin B = c/sin C
Law of Cosines
a2 = b2 + c2 – 2bc(cosA)
b2 = a2 + c2 – 2ac(cosB)
c2 = a2 + b2 – 2ab(cosC)
VECTOR OPERATION
Vector A and its negative counterpart
Scalar Multiplication and Division
Vector Addition
Parallelogram Law
Triangle construction
Vector Subtraction
or
A-B
R’ = A – B = A + (-B)
Resolution of Vector
Extend parallel lines from the head
of R to form components
Sample Problem 2.1
Q = 60N
P = 40N
The two forces act on bolt at A
Determine their resultant
Solution
Q = 60N
P = 40N
Solution
Q = 60N
P = 40N
Solution
Q = 60N
P = 40N
Q = 60N
P = 40N
c
A
B
Law of Cosines
a2 = b2 + c2 – 2bc(cosA)
a
b
C
b2 = a2 + b2 – 2ac(cosB)
c2 = a2 + b2 – 2ab(cosC)
Solution
Q = 60N
P = 40N
Q = 60N
P = 40N
Law of Sinus
a/sin A = b/sin B = c/sin C
Solution
Q = 60N
Alternative Trigonometric Solution
We construct the right triangle BCD and compute
P = 40N
CD = (60N) sin 25˚ = 25.36 N
BD = (60N) cos 25˚ = 54.38 N
Q = 60N
tan A =
25.36 N
94.38 N
A=15.04˚
Sudut A
R=
25.36
R=97.73N
sin A
α = 20º + A = 35.04˚
R = 97.7 N , 35.0˚
Sample Problem No 2.2
SP 2.2
A barge is pulled by two tugboats.
If the resultant of the forces exerted by the tugboats is a 25 kN force
directed along the axis of the barge,
determine
(a) the tension in each rope given α = 45,
(b) the value of α for which the tension in rope 2 is minimum.
Solution
25kN
Tension for Graphical Solution
The parallelogram law is used; the diagonal (resultant) is known to be equal 25kN
and to be directed to the right. This sides are drawn paralled to the ropes.If the
drawing is done to scale, we measure
T1 = 18.5 kN
T2= 13.0 kN
25kN
25kN
18.30kN
12.94kN
25kN
25kN
25kN
12.5kN
25kN
21.65kN
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2.10 To steady a sign as it is being lowered, two cables are
attached to the sign at A. Using trigonometry and knowing that
the magnitude of P is 300 N, determine
a) The required angle if the responding R of the two forces
applied at A is to be vertical
b) The corresponding magnitude of R.
Solution (a)
360 N
35˚
R
300 N
Using the triangle rule and the law of sines
Solution (b)
360 N
35˚
R
300 N
We need to calculate the
value of βto find the value
of R
Law of Sinus
β= ?
a/sin A = b/sin B = c/sin C
R = 513 N
Rectangular Components of a
Force
• x- y components
• Perpendicular to each other
•
F = Fx i + Fy j
y
Fy = Fy j
•
F
j
•
O
θ
i
Fx = F cos θ
x
Fx = Fx i
Fy = F sin θ
Example 1
A force of 800N is exerted on a bold A as shown
in the diagram.
Determine the horizontal and vertical components
of the force.
Fx = -F cosα= -(800 N ) cos 35º = -655N
Fy = +F sinα= +(800 N ) sin 35º = +459N
The vector components of F are thus
Fx = -(655 N) i
Fy = +(495 N) j
And we may write F in the form
F = - (655 N) i + (459 N) j
Example 2
A man pulls with a force of 300N on a rope attached
to a building as shown in the picture.
What are the horizontal and vertical components of the
force exerted by the rope at point A?
Fx = + (300N) cos α
Fy = - (300N) sin α
Observing that AB = 10m
(a)
cos α = 8m = 8m = 4
AB 10m
5
sin α = 6m = 6m =
AB 10m
3
5
We thus obtain
Fx = + (300N) 4 = +240 N
5
Fy = - (300N) 3 = -180 N
5
We write
F = + (240N) i - (180 N) j
tan θ= Fx
Fy
(b)
F=
(2.9)
6.7
Example 3
A force F = (3.1 kN)i + (6.7 kN)j is applied to bolt A.
Determine the magnitude of the force and the angle θit forms with the horizontal.
Solution
First we draw the diagram showing the two
rectangular components of the force and the angle θ.
From Eq.(2.9), we write
6.7
θ= 65.17˚
Using the formula given before,
Fx = F cos θ
Fy = F sin θ
Addition of Cartesian Vectors
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R=P+Q
Rxi + Ryj = Pxi + Pyj + Qxi + Qyj
Rx= Px+ Qx
Ry= Py + Qy
Rx = Σ Fx
Ry = Σ Fy
Py j
Syj
P
Ryi
R
S
Pxi
Sxi
O
Qyj
Qxi
Q
O
Rxi
2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS
How we calculate the force ?
2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS
SAMPLE PROBLEM 2.3
F2 = 80 N
F1 = 150 N
F4 = 100 N
F3 = 110 N
Four forces act on A as shown.
Determine the resultant of the forces on the bolt.
Solution
( F2 cos 20˚ )
j
F2 = 80 N
( F1 cos 30˚ )
F1 = 150 N
- ( F2 sin 20˚ ) i
F4 = 100 N
F3 = 110 N
( F1 sin 30˚ ) j
i
( F4 cos 15˚ )
- ( F4 sin 15˚ i) j
- F3 j
Solution
( F2 cos 20˚ )
j
F2 = 80 N
( F1 cos 30˚ )
F1 = 150 N
- ( F2 sin 20˚ ) i
F4 = 100 N
F3 = 110 N
( F1 sin 30˚ ) j
i
( F4 cos 15˚ )
- ( F4 sin 15˚ i) j
- F3 j
( F2 cos 20˚ )
j
( F1 sin 30˚ ) j
( F1 cos 30˚ )
- ( F2 sin 20˚ ) i
i
( F4 cos 15˚ )
- ( F4 sin 15˚ i) j
- F3 j
Ry = (14.3 N) j
Rx = (199.1 N) i
PROBLEMS
2.21
Determine the x and y component of each
of the forces shown.
Solution
Solution
Problems 2.23
Determine the x and y components of each of the forces shown.
1.2 m
1.4 m
2.3 m
900N
950N
1800 N
1.5 m
2.0 m
Solution
2.59m
2.5 m
2.69m
Solution
2.59m
2.5 m
2.69m
2.9 EQUILIBRIUM OF A PARTICLE
• The resultant force acting on a particle is zero
•
R = ΣF = 0
•
ΣFx = 0
ΣFy = 0
20 N
A
20 N
The forces of 20 N acting on the line but
in opposite direction, passing through
point A having the same magnitude. This
produces a resultant of R = 0.
Equilibrium of Forces
F4 = 400 N
O
F1 = 300
30
O
F2 = 173.2
F1 = 300
F4 = 400 N
F3 = 200 N
F3 = 200
30
F2 = 173.2 N
An equilibrium system of forces
produces a closed force polygon
Example
Figure 2.27 shows four forces acting on A.
In figure 2.28, the resultant of the given forces
is determined by the polygon rule. Starting from point
O with F1 and arranging the forces in tip-to-tail fashion,
we find that the tip of F4 coincides with the
starting point O.
Thus the resultant R of the given system of forces is zero,
and the particle is in equilibrium.
The closed polygon drawn in Fig 2.28 provides a
graphical expression of the equilibrium of A.
To express algebraically the condition for the
equilibrium of a particle, we write
Resolving each force F into rectangular components,
we have
We conclude that the necessary and sufficient
conditions for the equilibrium of a particle are
Returning to the particle shown in figure 2.27,
we check that the equilibrium conditions are
satisfied. We write
Newton’s First Law of Motion
If the resultant force acting on a particle is zero,
the particle will remain at rest or
will move with constant speed in a straight line.
Sample Problem 2.4
In a ship-unloading operation, a 15.6 kN automobile is supported by a
cable. A rope is tied to the cable at A and pulled in order to centre the
automobile over its intended position. The angle between the cable
and the vertical is 2, while the angle between the rope and the
horizontal is 30. What is the tension in the rope?
TAB
15.6kN
TAC
15.6kN
TAC
TAB
Sample problem 2.6
2.13m
0.46m
180N
1.22m
270N
1.22m
As part of the design of a new sailboat, it is desired to determine the drag force which
may be expected at a given speed.
To do so, a model of the proposed hull is placed in a test channel and three cables
are used to keeps its bow on the centerline of the channel.
Dynamometer reading indicate that for a given speed, the tension is 180N in cable AB
and 270N in cable AE.
Determine the drag force exerted on the hull and the tension in cable AC.
Solution
Determine of the Angles
2.13m
0.46m
180N
First the angles α and βdefining the
direction of cables AB and AC are determined.
1.22m
1.22m
270N
α= 60.26˚
β= 20.56˚
FD
TAE = 270 N
0.46m
= 0.375
1.22m
β = 20.56˚
tanβ =
TAC
TAB = 180 N
2.13m
= 1.75
1.22m
α = 60.26˚
tanα =
Free body diagram
Choosing the hull as a free body, we draw the
free-body diagram shown. It includes the forces
exerted by the three cables an the hull, as well as
the drag force FD exerted by the flow.
Solution
2.13m
0.46m
Equilibrium Condition
180N
1.22m
1.22m
270N
R
TAC
α= 60.26˚
TAB = 180 N
β= 20.56˚
FD
TAE = 270 N
We express that the hull is in
the equilibriumby writing that the
resultant of all forces is zero
=
TAB + TAC + TAE + FD
α= 60.26˚
TAB = 180 N
TAC
Since more than three forces are involved,
we resolved the forces into X and Y components
β= 20.56˚
TAB = -(180N) sin 60.26˚ i + (180N) cos 60.26˚ j
FD
= -(156.3N) i
TAC = TAC sin 20.56˚ i
= TAC 0.3512 i
TAE = 270 N
+ (89.3N) j
+ TAC cos 20.56˚ j
+ TAC 0.9363 j
TAE = - (270N) j
Y
FD = FD i
(180N) cos
cos 60.26˚
60.26˚ jj
(180N)
X
- (180N) sin 60.26˚ i
from,
- (270N) j
R
=
TAB + TAC + TAE + FD
(-156.3N + TAC 0.3512 + FD ) i + (89.3N + TAC 0.9363 – 270 N) j = 0
TAB = -(180N) sin 60.26˚ i + (180N) cos 60.26˚ j
= -(156.3N) i
TAC = TAC sin 20.56˚ i
= TAC 0.3512 i
+ (89.3N) j
+ TAC cos 20.56˚ j
+ TAC 0.9363 j
TAE = - (270N) j
FD = FD i
from,
R
=
TAB + TAC + TAE + FD
(-156.3N + TAC 0.3512 + FD ) i + (89.3N + TAC 0.9363 – 270 N) j = 0
: - 156.3N + TAC 0.3512 + FD
=0
89.3N + TAC 0.9363 – 270 N = 0
TAC = 193 N
FD = 88.5 N
THE END