Vector lesson and problems

ENGT 215 - Statics

Chapter 2 – Resultant of Coplanar
Force Systems

Chapter Goals
 Force System – any set of forces treated as a group.
 Equivalent Force Systems – any two systems of forces which
have the same mechanical effect on a body.
 Resultant – a single force that is equivalent to a given force
system (same mechanical affect on the body).

Force System Resultant

 The goal of this chapter is to learn about equivalent
force systems and how to calculate the resultants for
planar force systems.

Vector Representation
Anatomy of a Force Vector:
 Represented by arrow AB showing the line of action.
 Length of AB represents force magnitude at some convenient scale.
 Direction for line of action is indicated by angle θ counter-clockwise
from positive “x” reference axis (Standard Position).
 Arrowhead indicates sense of force.
 Reference coordinate system (x-y axes) is established at the point
of application A.
Tip
B

F
Tail
A
Point of application

Vector Representation
 Equal
Vectors - Two vectors are equal if they have the same
magnitude AND the same direction as in (a). The line of action
may be different as shown.

 NegativeVectors - Two vectors are negatives of each other if
they have the same magnitude and opposite directions as in (b).

Resultant of Concurrent Forces
Adding Vectors – Finding Resultants:
 The resultant of multiple vectors is the vector sum of those
vectors.

 Vectors
cannot be added algebraically, they must be added
geometrically!

 Two Parallelogram Rulefor
 common methods adding vectors (finding the resultant);
Parallelogram - a planar four sided
figure with parallel opposite sides.

 Triangle Rule

Parallelogram Rule - Method for Adding Vectors :
 Recreate the two vectors to be added such that their tails are coincident.
The direction and magnitude of each vector should not change!
 Create a parallelogram by drawing two lines, one from the tip of each
vector and parallel to the other vector.
 The intersection of these lines is the tip of the resultant vector R = v + w.
 Create the resultant by drawing a new vector R.
 Tail of R should correspond with tail of the other vectors v and w.

 Tip of R will be where construction lines intersect.

*Note –
magnitude of R
is not the sum
of the
magnitudes of v
and w.


Parallelogram Rule:
 Propertiesof parallelograms useful for calculating force resultants:
 Sum of the interior angles (2B + 2C) is 360 o.

 Opposite sides are equal in length.

 ∠A + ∠B = 180º

 ∠A = ∠C

C B

A B C

Triangle Rule - Method for Adding Vectors:
 Recreate the two vectors such that the tail of second vector w is
coincident with the tip of the first vector v. The direction and
magnitude of each vector should not change!

 Createthe resultant by drawing a new vector R.
 The tail of R should correspond with tail of first vector v.

 The tip of R should correspond with tip of second vector w.

 Force Triangle – triangle formed by forces.


Polygon Rule: (Polygon – any closed figure with straight sides)
 An extension of the Triangle Rule.
 Sum of 3 or more coplanar vectors can be accomplished by
adding 2 vectors successively, forming a polygon.

Example: R = a + b + c  R = (a + b) + c

 Force Polygon – polygon formed by forces.
b

R = (a + b) + c a a +b c

R

Polygon Rule:
 A more general form of Polygon Rule for adding vectors:
 Recreate all summed vectors tip-to-tail.

 Create the resultant vector R.


The tail of R should correspond with the tail of the first
vector v1.

The tip of R should correspond with the tip of the last
vector v6.


 VectorSums are Commutative:
 Get the same result using (v + w) or (w + v).

 Applies to all methods for adding vectors.

R=w+v

R

v
w


Subtracting Vectors:
A vector can be subtracted by adding its negative.
 Remember, the negative of a vector has the same
magnitude but the opposite direction.

R = v + (-w)

Subtracting Vectors

Resultant: a system of concurrent coplanar forces acting on a
rigid body may be replaced by a single Resultant force which
equals the vector sum of the given forces.
The vector sums can be determined by two methods:
 Graphical Method –
 Use linear scale and protractor to lay out vectors as

previously described.
 Measure length and angle of resultant.

 Too Inaccurate! Don’t Use!

 Trigonometric Method –
 Lay out vectors as described but not necessarily to scale.

 Label the known angles and lengths of sides (magnitudes).

 Compute the resultant using trigonometry.


Law of Sines

Law of Cosines
 This is the method to be used in this class!

Example 1:
 Determine the resultant of the two forces F 1 and F2
acting on the hook.

F1 = 54N

60o

F2 = 60N

Example 1: Solution
 Determine the resultant of the two forces F1 and F1 = 54N
F2 acting on the hook. 60o
1. Draw the vector diagram.
F2 = 60N
2. Use the Law of Cosines to find magnitude R.
3. Use the Law of Sines to find direction of R.

R 2 = F12 + F22 − 2 F1 F2 cos φ Law of Cosines
R 2 = 60 2 + 54 2 − 2 ⋅ 60 ⋅ 54 ⋅ cos(120)
R = 98.77 ≈ 98.8 N
R = 98.8 N ∠ 28.3°

sin β sin 120
=  Law of Sines R F1 = 54N
F1 R
120o
F1 β ϕ 60o
sin β = sin 120
R F2 = 60N
β = 28.259o

Example 2:

 Given: when θ = 30º, the resultant of F1 and F2 is 1 kN in the
vertical downward direction.
 Find: the value of F1 and F2.

Ropes support an I-beam
of weight 1 kN. Thus
resultant is 1 kN vertically
downward.

Example 2: Solution

 Given: when θ = 30º, the resultant of F1 and F2 is
1 kN in the vertical downward direction.
 Find: the value of F1 and F2.

Solution:

1. Draw the vector diagram
2. Use the sine law to find
F1 and F2

F1 1000
o
= o
∠
⇒ F1 = 653 N 290º
Sin30 Sin130
F2 1000
o
= o
⇒ F2 = 446 N 240º
∠
Sin 20 Sin130

Exercise 1:
 Determine the magnitude and direction of the resultant
of the two forces acting on the eye hook.
 Hint: The law of cosines is handy.

Exercise 1: Solution
 Determine the magnitude and
direction of the resultant of the two
forces acting on the eye hook.
Magnitude by Law of Cosines:
R2 = F12 + F22 – 2 F1F2cos 30
Y
R2 = 852 + 1502 – 2 (85)(150)cos 30
F2 = 150 N
R2 = 7641.4 N2
R= 87.4 N R
θ 30º X
Angle by Law of Cosines: F1 = 85 N
F22 = F12 + R2 – 2 F1Rcosθ
cosθ = (F12 + R2 – F22 ) / (2F1R) *Note: to determine
an angle > 90º such
cosθ = (852 + 87.42 – 1502 ) / (2*85*87.4) as θ, use the law of
cosθ = -0.5139 Cosines instead of
θ = 120.9º R = 87.4 N ∠ 120.9° the law of Sines.

Exercise 2:
 Determine the magnitude and direction of the resultant
of the two forces acting on the eye hook.

 Determine the magnitude and
direction of the resultant of the two
forces acting on the eye hook.
Magnitude by Law of Cosines: Y
R2 = F12 + F22 – 2 F1F2cos 35 F2 =
65 N
R2 = 802 + 652 – 2 (80)(65)cos 35 R
R2 = 2105.82 N2 25°
R= 45.8892 = 45.9 N α 10°
F 1 = 80
N
θ 10° X
Angle by Law of Sines: 30º
(sin α / 65 N) = (sin 35° / 45.8892 N)
sin α = (65)(sin 35°) / 45.8892 N *Note: to determine
α = 54.3355° an angle > 90º such
as θ, use the law of
θ = α + 10º Cosines instead of
θ = 64.3º R = 45.9 N ∠ 64.3° the law of Sines.

Rectangular Components
 Force Component: A resultant is a single force that is equivalent to
the sum of a group of forces. In other words it can replace all those
forces and give the same effect. The individual forces that the
resultant replaces are called the components of the resultant.

F1 F Example: F1 and F2 are
components of resultant F
F2
 Rectangular Components: 2 mutually perpendicular components.
 Typically selected along horizontal x and vertical y axes.

 Think of F and F as the projection of F onto the X and Y axes.
x y
y
Example: Fx and Fy are
Fy
F rectangular components of F
θ x
0 Fx + Fy = F
Fx

Rectangular Components:
 Rectangular components are useful because they isolate the effect
of the resultant force in each direction.
 The effect of the force in the X direction acts independently of the
force effect in the Y direction (and vice versa).
 Using Rectangular Components, the effect of the resultant force in
each direction can be treated independently.

Example: Non-rectangular components
F1 F
F1 and F2 both have a force effect in
the X direction.
F2
y
Example: Rectangular components
F Only Fy has an effect in Y direction and
Fy
θ x only Fx has an effect in the X direction.
0 Fx They can be treated independently.

 Magnitude of rectangular components: If the magnitude and
direction of F is known, the magnitudes of the rectangular
components can be calculated from right triangle trig.
Magnitude
Fx = F cos θ Fx is called the x-component of F.
Fy = F sin θ Fy is called the y-component of F.
y y

Fy F F
Fy
θ x θ x
0 0
Fx Fx

Rectangular Components y
 Magnitude of Rectangular Components: F Fy
What if F is directed as shown? θ x
Fx 0
 Assume Standard Position:  Use Reference Angle (α):
y y

Fy F
F
Fy
θ α θ x
x
0 0 Fx
Fx
Angle θ measured from +x axis. Use α (positive acute angle from +x
Equations yield direction sign of or –x axis).
component. Direction sign found by inspection.

Example: if θ = 150° Example: if θ = 150°, α = 30°
Fx = F cos 150 = -0.866 F Fx = F cos 30 = 0.866 F  -0.866 F
Fy = F sin 150 = 0.500 F Fy = F sin 30 = 0.500 F

Rectangular Components – Reference Angle
For any angle θ in standard position, the Reference Angle α
(pronounced “alpha”) is defined as the acute positive angle
between the vector and the x-axis.

•Reference Angle α is acute (0º < α < 90º).
•Reference Angle α is always positive.

II
To Calculate Reference Angle α I
I 0º < θ < 90º, then α = θ
II 90º < θ < 180º, then α = 180º - θ
III 180º < θ < 270º, then α = θ - 180º
IV 270º < θ < 360º, then α = 360º - θ
III IV

Example 3:
 Resolve the 60 N force on the ring into its horizontal (x)
and vertical (y) components.

60°

F = 60N

y
Example 3: Solution Fx
 Resolve x
the 60 N force on the ring into its
horizontal and vertical components. 60°
Method 1: Standard Position
Given: F = 60N, α = 60° Fy
Find: Fx, Fy F = 60N
y Method 2: Reference Angle
Solution:
Fx = F cos θ x Given: F = 60N, α = 60°
60° Find: Fx, Fy
θ = 360° - α
θ = 360° – 60° F = 60N Solution:
θ = 300° Fx = F cos α →
Fx = 60 cos 300° N Fx = 60 cos 60° N
Fx = 30.0 N Fx = 30.0 N →
Fy = F sin θ Fy = F sin α ↓
Fy = 60 sin 300°N Fy = 60 sin 60° N
Fy = -51.96 Fy = 51.96 N ↓
Fy = -52.0 N Fy = 52.0 N ↓

Example 4:
 Resolve
the weight of the 150 lb skater into
components along the rail and normal to the rail.

1 ft

2 ft

W = 150 lb

Example 4: Solution
 Resolve
the weight of the 150 lb skater into
components along the rail and normal to the rail.
1 ft
Given: W = 150 lb, rail slope= 1:2. 2 ft
Find: Wx, Wy W = 150 lb
Solution: x
tan α = 2 ft / 1 ft y
α = tan-1(2 / 1) α 1 ft
α = 63.43°
2 ft
Wx = W cos α  Wx
α
Wx = 150 cos 63.434 lb 
Wy
Wx = 67.1 lb  or Wx = -67.1 lb
Wy = W sin α 
W = 150 lb
Wy = 150 sin 63.434° lb 
Wy = 134.2 lb or Wy = -134.2 lb

Exercise 3:
 Resolve the 75 N force on the ring into its horizontal
and vertical components.

48°

F = 75N

 Resolve the 75 N force on the ring into its
horizontal and vertical components. 48°

F = 75N
Method 2: Reference Angle
Given: F = 75 N, α = 48° y
Find: Fx, Fy
Solution: Fx x
Fx = F cos α ←
48°
Fx = 75 cos 48° N
Fx = 50.2 N ← or -50.2 Fy
Fy = F sin α ↓
Fy = 75 sin 48° N F = 75N
Fy = 55.7 N ↓ or -55.7

Rectangular Components - If the magnitudes of Fx and Fy are known,
the magnitude and direction of F can be calculated. y
 Magnitude of F:
F = (Fx2 + Fy2)½ F
Fy
 This is the Pythagorean Theorem α θ x
Fx 0
F2 = Fx2 + Fy2
 Direction of F: α = tan-1Fy / Fx Given: Fx = -20 N, Fy = 5N
Find: F
 α is the angle of vector F with Magnitude F = (Fx2 + Fy2)½
respect to x axis (pos. or neg.). F = (400 + 25)½ = 20.6 N
 To find the direction angle θ, must Direction α = tan-1Fy / Fx
know which quadrant vector lies in. α = tan-15 / 20= tan-1(0.25)
α = 14.04°
I 0º < θ < 90º, then θ = α
II 90º < θ < 180º, then θ = 180º - α 2nd quadrant since Fx is neg.
III 180º < θ < 270º, then θ = 180º + α θ = 180 -14.04 = 166.0°
IV 270º < θ < 360º, then θ = 360º - α θ = 166.0°
or θ = - α F = 20.6 N ∠ 166.0°

Example 5:

 Therectangular components of a force are given as Fx = -450 N
and Fy = -300 N. Find the magnitude and direction of the force.

y

Fx= -450 N x
0

Fy= -300 N

Example 5: Solution y
 Therectangular components of a force are given
as Fx = -450 N and Fy = -300 N. Find the Fx= -450 N
magnitude and direction of the force. x
0
Given: Fx = -450 N, Fy = -300 N y
Find: F Fy= -300 N
Magnitude –
F = (Fx2 + Fy2)½ Fx= -450 N θ x
F = [(-450) + (-300) ]
2 2 ½
α 0
F = 540.83 N F
F = 541 N
Direction- Fy= -300 N
α = tan-1Fy / Fx
α = tan-1300 / 450= tan-1(0.667)
α = 33.7°
3rd quadrant since Fx and Fy are neg.
I 0º < θ < 90º, then θ = α
θ = 180 + α II 90º < θ < 180º, then θ = 180º - α
θ = 180 + 33.7° III 180º < θ < 270º, then θ = 180º + α
θ = 214° IV 270º < θ < 360º, then θ = 360º - α
F = 541 N ∠ 214° or θ = - α

Exercise 4:

 Therectangular components of a force are given as Fx = 125 N
and Fy = -288 N. Find the magnitude and direction of the force.

y

Fx= 125 N
x

Fy= -288 N

 The y
rectangular components of a force are given
Fx= 125 N
as Fx = 125 N and Fy = -288 N. Find the
magnitude and direction of the force. x
y Fy= -288 N
Given: Fx = 125 N, Fy = -288 N
Find: F
Magnitude –
F = (Fx2 + Fy2)½ Fx= 125 N
θ
F = [(125) + (-288) ]
2 2 ½
0 x
α
F = 313.96 N
F = 314 N
Fy= -288 N F
Direction-
α = tan-1Fy / Fx
α = tan-1288 / 125= tan-1(2.304)
α = 66.5°
2nd quadrant since Fy is neg.
I 0º < θ < 90º, then θ = α
θ=-α II 90º < θ < 180º, then θ = 180º - α
θ = -66.5° III 180º < θ < 270º, then θ = 180º + α
F = 314 N ∠ -66.5° or 314 N ∠ 293.5° IV 270º < θ < 360º, then θ = 360º - α
or θ = - α

Resultants by Rectangular Components
Resultants by rectangular components: The resultant of any number
of concurrent coplanar forces can be calculated by resolving each
into its rectangular components. y
 Step 1: Resolve each force into its F1
F2
rectangular components.
x
 Step 2: All x components are directed 0
horizontally and can be added algebraically F3
to get x component of resultant (Rx).
Rx = ΣFx = (F1)x+ (F2)x + (F3)x +…
 Step 3: All y components are directed vertically and can be added
algebraically to get y component of resultant (Ry).
Ry = ΣFy = (F1)y+ (F2)y + (F3)y +…
 Step 4: Using x and y components of resultant calculate magnitude
and direction of resultant R.
R = (Rx2 + Ry2)½ α = tan-1Ry / Rx

Example 6:
 Determine the resultant of the two forces using
rectangular components;
F1 = 3 kN ∠ 32° and F2 = 1.8 kN ∠ 105°.

y

F2 F1

x
0

y
Example 6: Solution F2
F1

 Determine the resultant of the two forces x
0
F1 = 3 kN ∠ 32° and F2 = 1.8 kN ∠ 105°.
Given: F1 = 3 kN ∠ 32° and F2 = 1.8 kN ∠ 105°.
Find: Resultant R y R
X Components F1 – Y Components F1 –
θ2
Fx1 = F1 cos θ1 Fy1 = F1 sin θ1 F2 F1
Fx1 = 3000 cos 32° N Fy1 = 3000 sin 32° N
Fx1 = 2544.1 N Fy1 = 1589.8 N θ1 x
X Components F2 – Y Components F2 – 0
Fx2 = F2 cos θ2 Fy2 = F2 sin θ2
Fx2 = 1800 cos 105° N Fy2 = 1800 sin 105° N Direction R –
F = -465.9 N F = 1738.7 N α = tan-1Ry / Rx
X x2
Components R – Yy2Components R –
α = tan-13328.5 / 2078.2
Rx = Fx1 + Fx2 Ry = Fy1 + Fy2
α = tan-1(1.602)
Rx = 2544.1 – 465.9 N Ry = 1589.8 + 1738.7 N α = 58.0°
Rx = 2078.2 N Ry = 3328.5 N Rx & Ry positive so 1st quadrant
Magnitude R – θ = α = 58.0°
R = (Rx2 + Ry2)½
R = 3920 kN∠ 58.0°
R = [(2078.2)2 + (3328.5)2]½ = 3923.9 N = 3920 N
[

Exercise 5A:
 Determine the resultant of the two forces using
F1 = 150 N ∠ 28° and F2 = 100 N ∠ 99°.

y

F2 F1

x
0

y
Exercise 5A - Solution F2
F1

 Determine the resultant of the two forces x
0
F1 = 150 N ∠ 28° and F2 = 100 N ∠ 99°.
Given: F1 = 150 N ∠ 28° and F2 = 100 kN ∠ 99°.
Find: Resultant R y R
X Components R – θ2
Rx = Fx1 + Fx2 F2 F1
Rx = F1 cos θ1+ F2 cos θ2
θ1 x
Rx = 150 cos 28° + 100 cos 99° N
0
Rx = 116.8 N
Y Components R –
Ry = Fy1 + Fy2 Direction R –
Ry = F1 sin θ1+ F2 sin θ2 α = tan-1Ry / Rx
Ry = 150 sin 28° + 100 sin 99° N α = tan-1169.2 / 116.8
Ry = 169.2 N α = 55.381°
Rx & Ry positive so 1st quadrant
Magnitude R –
R = (Rx2 + Ry2)½ θ = α = 55.4°
R = [(116.8)2 + (169.2)2]½ = 206 N
[ R = 206 N∠ 55.4°

Exercise 5B:
 Determine the resultant of the three forces using
F1 = 100 N, F2 = 200 N, and F3 = 300N.
y

F2 40°
F1
20° x
0
60°

F3

Moment of a Force
What is the Moment of a Force?
 A force can have two effects on a rigid body;
 Translation - tends to move it linearly along its line of action.

 Rotation - tends to rotate it about an axis.

 Moment of a Force: the measure of a force’s tendency to rotate
the body about an axis.
 The moment of a force
(moment) is also called torque
(Ex: torque wrench).

Moment of a Force
What is the Moment of a Force?
 Moment of a Force in Action: Whenever there is a tendency for
rotational motion to occur, a Moment is at work.

What factors would
increase the tendency
for rotation to occur in
the following examples?

Moment of a Force
 Moment of a force : The tendency of a force to cause rotation
(moment of the force) depends on two factors;
 The magnitude of the force.

 The perpendicular distance (d) from the center of rotation (O)

to the line of action of the force.
 An important distinction:
 Distance (d) - perpendicular distance from the center of

rotation (O) to the line of action.
 Distance (L) - distance from the center of rotation (O) to the

point of force application.

L

L

Moment of a Force
 Mathematical Definition of a Moment: The moment Mo of a force
F about a point O is equal to the magnitude of the force
multiplied by the perpendicular distance d from O to the line of
action of the force.
Mo = F·d

 Point O is referred to as the Moment Center
 Distance d is referred to as the Moment Arm.

 Units- in terms of force & distance. L
 SI Units: N·m or kN·m

 English Units: ft·lb or in·lb O

Moment of a Force

A
Let’s test the
mathematical
definition of a moment
and see its affects
using your text book!

B

Moment of a Force
 Direction of a Moment: A moment is a vector just like a force is a
vector. The direction of a moment vector is determined by the
right hand rule.

 PositiveMoment - the force tends to cause a
counter-clockwise rotation about the moment
center .
 Negative Moment - the force tends to cause a
clockwise rotation about the moment center .
Positive Moment

Moment of a Force

 Summation of Moments: In the 2D case (planar forces),
moments can be added algebraically just like forces with
the same line of action (Rx = F1x + F2x+ …).
 Counter-clockwise  moments positive.

 Clockwise  moments negative.

y
F1

d1
Mo = Mo1 + (-Mo2) + (-Mo3) +…

d2
o x
F2 d3

F3

Moment of a Force
 Two ways to calculate the moment of a force about a point;
Transmissibility Method: Rectangular Component Method:
 Extend the line of action to Find the component of the force
find the moment arm length d perpendicular to rod L.
(⊥ distance from center to line  Calculate the moment using;
of action). Mo = Fx·d where
 Calculate the moment using;
Fx is the normal component of F
Mo = F·d where
d is the moment arm = L y
d is the moment arm ≠ L

ϕF Fx
ϕ
x
L

L
d=
F
d

o o

Example 7:
 Determine the moment of a 500 N force about
point o, if θ = 30°, 60°, 90°, and 120°.

θ F
m
m
0
20

60°
o

Example 7: Solution
 Determine the moment of a  θ = 60 °
500 N force about point o, if sin θ = d / .200 m
θ = 30°, 60°, 90°, and 120°. d = .200 sin 60° mm
θ F d = 0.1732 m

m
Mo= Fd

m
0
Mo= - (500 N)(0.1732 m)

20
 θ = 30 °
sin θ = d / .200 m Mo= -86.6 Nm or
d = .200 sin 30° m 60° Mo= 86.6 Nm 
d = .100 m o  θ = 90 °
Mo= Fd sin θ = d / 200 mm
θF
Mo= - (500 N)(0.1 m) d = 200 sin 90° mm
m
m

Mo= -50 Nm or d = 200 mm = 0.2 m
0
20

Mo= 50 Nm  Mo= Fd
d Mo= - (500 N)(0.2 m)
o Mo= - 100 Nm or
Mo= 100 Nm

Example 7: Solution
 Determine
θ F
the moment of a 500 N force

m
m
about point o, if θ = 30°, 60°, 90°, and

0
20
120°.
F
60°
ϕ θ o
 θ = 120 °
d ϕ = 180° - θ
m
m

ϕ = 180° - 120° = 60°
0
20

sin ϕ = d / 200 mm
o d = 200 sin 60° mm
d = 173.2 mm = 0.1732 m
Mo= Fd
Mo= - (500 N)(0.1732 m)
Note: Same value Mo= - 86.6 Nm or
as for θ = 60º
Mo= 86.6 Nm

Exercise 6:
 Determine the moment of force F= 150 lb about
point o given the rod length (L) is 16 inches.

o

L = 16 in
θ=128°

F= 150 lb

 Determine the moment of force F= 150 lb about
point O given the rod length (L) is 16 inches.
Given: F = 150 lb
θ = 128°
o
L = 16 in
Find: Mo
L = 16 in Solution:
d
θ=128° ϕ = 180° - 128° = 52°
ϕ d = L sin ϕ = 16 sin 52° in
F= 150 lb d = 12.6082 in
Mo = Fd = (150 lb)(12.61 in)
Mo = 1891 in lb 
Mo = (1891 in lb)(1 ft / 12 in)
Mo = 157.6 ft lb 

Moment of a Force
 Varignon’s Theorem: The
moment of a force about
any point is equal to the
sum of the moments
produced by the
components of the force
about the same point.
y
 Thisis the principle used in
Fy Fx this method for moment
calculation. In this case, the
ϕ
L

moment due to Fy is zero.
d=

x
F

*For Moment calculations its important to
remember Transmissibility Law – Force
can be moved along line of action only.
o Don’t move Fy tip to tail with Fx.

Exercise 7:

 By Varignon’s Theorem, force F can be broken into
its components acting at point A. Using this approach,
find the moment about O due to the 250 N force.

y

A 135°
x

100 mm F
o
B
200 mm

Exercise 7: Solution 1
 Determine the moment of the 250 N force about point O.

y
Given: F = 250 N; ∠ = 135°
LAB= 100 mm; LBO= 200 mm
135° Find: Moment Mo about point O.
A Fx x Solution:
α
α = 135°-90° = 45°
100 mm Fy Fx = Fcos α = 250N (cos45°)
F o Fx = 176.7767N
B Fy = Fsin α = 250N (sin45°)
200 mm Fy = 176.7767N
Mo = (0.2m)Fy – (0.1m)Fx
Mo = (0.1m) 176.7767N
Mo= 17.68 Nm 

Exercise 7: Solution 2
Given: F = 250 N; θ = 135°;
y LAB= 100 mm; LBO= 200 mm
Find: Moment Mo about point O.
Solution:
135°
A Tan (45° +α) = LBO / LAB = 0.2 / 0.1
Fy Fx Tan (45° +α) = 2.0
100 mm 45° L (45° +α) = 63.4349°
α OA
F o α = 18.4349°
B x F = Fsin α = 250 sin 18.4349° N
y
200 mm Fy = 79.0567 N
Let d = L = LOA
d = (0.22 + 0.12)½ = 0.22361 m
Mo = Fyd = (79.0567 N)(0.2236 m)
(
Mo = 17.68 Nm 

Moment of a Force
 Transmissibility (review): The point of application of a
force acting on a rigid body may be placed anywhere
along its line of action.
 Example below shows how this is useful.

Example 8:

 Use the geometry method in which d (moment arm
perpendicular to line of action) is determined.

A 135°

100 mm F
o
B
200 mm

Example 8: Solution
 Determine the moment
of the 250 N force about Given: F = 250 N; θ = 135°
point O. LAB= 100 mm; LBO= 200 mm
Find: Moment Mo about point O.
Solution:
ϕ = 180° - 135° = 45°
A 135° tan ϕ = (LBC / 100) mm
LBC = (tan ϕ / 100) mm
100 mm ϕ F LBC = tan 45° / 100 = 100 mm
α o LCO = 200 – LBC = 100 mm
B C α
d α = 90° - ϕ = 45°
200 mm d = LCO sin α = 100 sin 45° mm
d = 70.71 mm = 0.07071 m
Mo = Fd = (250 N)(0.07071 m)
Mo = 17.68 Nm 

Exercise 8:

 Considering Varignon’s Theorem and
Transmissibility, find the moment about O
due to the 250 N force.

y

A 135°
x

100 mm F
o
B
200 mm

Force Couples
 Force Couple: A couple is defined as two parallel forces with
equal magnitudes but opposite sense having different lines of
action separated by a distance (d).
 Net force of a couple is zero, but rotates in specified direction.

 Moment of a couple is not zero.

 Causes rotation about an axis perpendicular to plane of its

forces.
 A force couple is a way to produce a moment without a net force.

Equivalent

M = Fd

Force Couples
 Moment of a Couple: The moment of a
couple is given by;
M = Fd

 The moment of a couple is independent of
the choice of the moment center.
MO = F (a+d) – Fa
MO = Fa + Fd – Fa
MO = Fd  thus MO is independent
of distance a.
a

 Themoment of a couple about any point,
anywhere, is the same. O

A couple may be transferred to any location
in its plane and still have the same effect.

Force Couples
 EquivalentCouples: Two couples acting in the same plane are
equivalent if they have the same moment acting in the same
direction  produce the same mechanical effect.

If M = M’ then these are Equivalent Couples, and
0.4F = 0.3F’  F = (¾)F’

Force Couples
 Addition of Couples: The moments about a point of two or
more couples acting in a plane may be added algebraically.
Using our convention ccw  = positive; cw  = negative.

 Example: Find the total
F2 moment about point P.
P F2 d2 Mp = -(F1d1) + -(F2d2)

What direction does the
F1 moment act?
d1
F1

Exercise 9:
 Captain Bligh encounters rough seas. Determine the
moment of the couple required to keep his ship on
track if F= 50 lb and the wheel diameter d = 30 inches.

-F F
d

 Captain Bligh encounters rough seas. Determine the
moment of the couple required to keep his ship on
track if F= 50 lb and the wheel diameter d = 30 inches.

Given: F = 50 lb
d = 30 in
Find: M
Solution:
M = F(d/2)  + F(d/2)  = Fd 
M = (50 lb)(30 in) 
M = 1500 in lb 
-F F
d

Exercise 10:
 Determine the moment of the force couple acting
on the block.
 Then determine the force value for an equivalent
couple having opposing forces acting horizontally
at C & D.
Fab = 40 N
C
w=120 mm B

h=90 mm
A D
Fab = 40 N

 Determine the moment of
Fab = 40 N the force couple acting on
C the block. Then replace with
w=120 mm B an equivalent couple acting
horizontally at C & D.
h=90 mm Given: Fab = 40 N; w = 120 mm;
A h = 90 mm
D
Fab = 40 N Find: M and equivalent couple at
C & D.
Equivalent Solution:
Couples M = Fabw 
Fcd C M = (40N)(120 mm) 
B
w=120 mm M = 4.80 Nm 
Fcdh = M = 4.80 Nm 
h=90 mm Fcd Fcd = M/h = (4.80 Nm)/(0.900 m)
A D Fcd = 53.3 N (directions as shown)

Non-Concurrent Force Systems

 In previous sections we;
 Found the resultant of concurrent force systems.

 Learned to calculate the moment of a force about a

point (axis).
 Learned about force couples and the moments they

produce.
 In this section we focus on;
 Non-Concurrent Force Systems .

 Use what we learned about moments to find the

resultant of a Non-Concurrent Force System.

Force - Couple Systems
A force system can be replaced by an equivalent system
consisting of a combination of forces and couples (moments).
 The net effect must be identical for each system – i.e. they

must both produce the same mechanical effect.

 Equivalent Force Systems – systems of forces are equivalent if;
they have the same resultant force, and
the same resultant moment about any selected point.

 Are these Equivalent force
systems? (transmissibility)

Are these Equivalent
force systems? 
(parallel force displacement)

Force - Couple Systems
 Force-Couple System – any force may be moved to another
point without changing its mechanical effect, provided that an
appropriate couple (moment) is added.
 The value of the added couple (moment) is equivalent to the
moment of the force at its original location about its new location.

F F
Not

Equivalent

d d d
F F F F F F

-F -F M=Fd
Proper Method: Step 1: Step 2: Step 3:

Example 9:

 Replace the 2 KN force F with a
force-couple system at point B.

A F = 2 KN

h=390 mm

B

w=1.4 m

Example 9: Solution
 Replace the 2 KN force F with a force-couple system at point B.
A
F= 2 KN
h=390 mm Given: F = 2KN
h = 390 mm
B
w = 1.4 m
w=1.4 m Find: Equivalent force-couple system
A at B.
Solution:
F=2 KN Step 1: Step 1: Move the force to point B.
B
Step 2: Calculate moment about B
due to F at A.
A M = F(0.390 m) 
M = (2000 N)(0.390 m) 
F=2 KN M = 780 Nm  or -780 Nm
B Step 3:
Step 3: Place moment at B.
M=780 Nm

Example 10:

 Replace the three forces shown with an
equivalent force-couple system at A.

F1
F2
F3

Example 10:
To find the equivalent
set of forces at A.
3
θ = tan   = 36.87 o
−1

4
Rx = ∑ Fx
= 400 N cos ( 180 ) + 750 N cos ( 36.87 ) + 100 N cos ( 90
o o o
)
= 200 N

Ry = ∑ Fy
= 400 N sin ( 180o ) + 750 N sin ( 36.87 o ) + 100 N sin ( 90 o )
= 550 N

Example 10:
Find the moments about point A.

Using the line of action for the force at B. The force
can be moved along the line of action until it reaches
perpendicular distance from A

uuu uur
r
M 1 = FB d
= 100 N ( 360 mm )
= 36000 N-mm

Example 10:

The force at O can be broken up into its two components
in the x and y direction

Fx = 750 N cos ( 36.87 o )
= 600 N
Fy = 750 N sin ( 36.87 o )
= 450 N

Using the line of action for each component, their
moment contribution can be determined.

Example 10:
Using the line of action for Fx component d is 160 mm.
uuu
r uuu r
M 2 = FOx d
= 600 N ( 160 mm )
= 96000 N-mm
Fy component is 0 since in
line with A.
uuur uuu
r
MB = ∑ Mi
uuu uuu uuu
r r r
= M1 + M 2 + M 3
r r r
= 36000 N-mm k + 96000 N-mm k + 0 N-mm k
r
= 132000 N-mm k

Example 10:

The final result is
R = 585 N at 70.0o
M = 132 Nm 
M = 132 Nm

Resultant of Nonconcurrent
Coplanar Force Systems
 Previouslywe found the resultant (single force equivalent to the
given force system) for a Concurrent, Coplanar force system – all
the forces passed through a single point.

 Ina Nononcurrent Colanar force system;
 No concurrent point exists.

 Line of action for resultant is not immediately known.

Q: If the forces are Nonconcurrent, how do we find the resultant?
A: We need to introduce moments into our resultant formulation.

 Remember! Equivalent Force Systems – systems of forces are
equivalent if;
they have the same resultant force, and
the same resultant moment about any selected point.

Steps to find the resultant of a Nonconcurrent Coplanar force system:

 Step1: Find the magnitude and direction of the resultant Force.
Same as before.
 Choose convenient x-y coordinate system.

 Resolve each force into x and y components.

 Components of the reaction force are the algebraic sums of the
individual force components;
Rx = ΣFx Ry = ΣFy

 Find magnitude of the resultant R = (Rx2+ Ry2)½

 Find the direction of the resultant α = tan-1|Ry / Rx|  then find θ

Steps to find the resultant of a Nonconcurrent Coplanar force system:

 Step 2: Find the location for the line of action of the resultant.
 The moment of the original force system and the moment of the

resultant force about an arbitrary point must be equal.

 Calculate moment of the original force system about any
convenient point.

 Knowing this and the resultant direction & magnitude, the
resultant location can be calculated.

Example 11:

 Find the resultant and its location for the 2 force system shown.

A F1= 9.32 KN

h=300 mm

B C

w=1.4 m F2= 10 KN
40º

Example 11: Solution

Solution:
A F= 9.32 KN Step 1: Magnitude & direction of
resultant.
h=300 mm
Break 10KN force into components;
C
B Fx = 10KN cos 40º Fy = 10KN sin 40º
w=1.4 m F= 10 KN Fx = 7.66 KN Fy = 6.43 KN
40º

A F= 9.32 KN Find Rx = ΣFx
h=300 mm Rx = (9.32 – 7.66) KN
C Rx = 1.66 KN or 1.66 KN 
B Fx= 7.66 KN
w=1.4 m
Find Ry = ΣFy
Fy= 6.43 KN
Ry = 0 + 6.43 KN
Ry = 6.43 KN or 6.43 KN

A
F= 9.32 KN
h=300 mm Solution:
C
B Fx= 7.66 KN Find magnitude of R:
w=1.4 m R = (Rx2+ Ry2)½
Fy= 6.43 KN R = [(1.662+ 6.432)]½ KN
R = 6.64 KN
R

Ry= 6.43 KN Find direction of R:
α = tan-1|6.43/1.66|
75.5º
α = 75.5º = θ
Rx= 1.66 KN
R = 6.64 KN ∠ 75.5º
R= 6.64 KN
75.5º

A
h=300 mm F= 9.32 KN Solution:
C Step 2: Find location of resultant.
B Fx= 6.13 KN Find moment of force system about C:
w=1.4 m
MC = ΣMF
Fy= 5.14 KN
MC = (9.32KN)(300mm) + 0 + 0 
y MC = 2.796 Nm  or -2.796 Nm

R must produce this same moment,
A so locate R accordingly. By inspection,
dy
R must be to the left of C to produce a
C negative moment . Draw line of
B x action and solve for dx and dy which
dx
give MC = 2.796 Nm  or -2.796 Nm
R
75.5º

y Review data: MC = 2.796 Nm 
Y intercept
Rx = 1.66 KN Ry = 6.43 KN
X intercept A At x intercept, Rx produces no moment
dy About C.
C Rydx = MC
B x
dx dx = MC / Ry= 2.796 Nm / 6.43 KN
R dx = 0.434 m
75.5º
At y intercept Ry produces no moment
y
Rxdy = MC
dy = MC / Rx= 2.796 Nm / 1.66 KN
A dy = 1.684 m dy = 1.684 m
θ=75.5º R
θ C
B x
Any location along line of action is
dx = 0.434 m acceptable. Choose x = -0.434 , y = 0
since located on part.

Exercise 11:
 Find
the resultant for the 3 force system
shown and locate it with respect to point A.

y

F2=200 N
F1=500 N
4 F3=450 N

B 3 C D
A
x
1.5 m 1.5 m 1m

 Find
the resultant for the 3 force system
shown and locate it with respect to point A.

y
Rx = -300 N
Ry = -650 N X = 2.77 m
R = 716 N 200 N
500 N
α= 65.2° 4 450 N
θ = 245° B 3 C D
R = 716 N ∠ 245° A
x
MA= -1800 Nm 1.5 m 1.5 m 1m
X = 2.77 m
R = 716 N ∠ 245°

Resultant of Distributed Line Loads
 So far we’ve dealt with loads developed
by individual forces concentrated at a
point called Point Loads or Concentrated
Loads.

 But what if a load is continuous along
the length of a beam, say due to sacks
of concrete.

 This continuous loading is referred to as a Distributed Load. It
may be exerted along a line, over an area, or throughout an
entire body (volume – remember gravity?). We will only
consider the line load.

 Goal: determine how to represent a distributed line load as a
point load resultant which we are more familiar with.

 LoadIntensity (w) – the magnitude of the load per unit of length
over which it acts. (lb/ft) (N/m)
example: 300 lb/ft - each foot of load represents 300 lb of force.

Types of Distributed Loads:

 Uniform Load – distributed load with
constant intensity (w).
example: 400 N/m uniform load
distributed over 2m equals 800 N total.

 Triangular Load – distributed load
whose intensity varies linearly from 0 to
some maximum value (wo).
example: 300 N/m triangular load
distributed over 2m equals 300 N total.


Types of Distributed Loads:

 Trapezoidal Load – distributed load
whose intensity varies linearly from a
non-zero value to some maximum
value. Can be treated as a uniform
load plus a triangular load.
example: a trapezoidal load with a minimum intensity of 200 N/m
and a maximum intensity of 300 N/m distributed over 2m equals
(200 N/m)(2 m) + (½)(100 N/m)(2m) = 500 N total.

 EquivalentConcentrated Force: to determine the resultant of a
force system involving distributed loads, each distributed load
may be replaced by its equivalent concentrated load as follows;

 Magnitude – equal to area of loading diagram.
 Direction – according to distributed load direction (typically

vertically down).
 Location – line of action passes through centroid of loading

diagram.

 For Trapezoidal Load – break into Uniform and Triangular load
first (treat them independently).

Exercises 12:
 Convert each distributed load to point load(s) of appropriate
magnitude and locate correctly with respect to left end of beam.
30 kN
6 kN/m

B 3 kip/ft
A 1 kip/ft

2m 3m 1m A B

9 ft

Law 10 kips 30 kN 5 kips
Rw 1 kip/ft
P1 P2
4 kip/ft La Lb

A B
A B
Ra A B
Rb
2m 9 ft 3m 61m
ft 10 ft Ra
9 ft
Rb
5 ft

Exercises 12: Solution
 Convert each distributed load to a point load of appropriate
magnitude and locate correctly on the beam.
30 kN
6 kN/m

(6 kN/m)(5m) = 30 kN at 2.5 m
A B

2m 3m 1m

3 kip/ft
1 kip/ft
(1 kip/ft)(9ft) = 9 kip at 4.5kN
Law 30
ft
A B
(½)(2 kip/ft)(9ft) = 9 kip at 6 ft
Rw
9 ft

A B
Ra 10 kips Rb 5 kips
2m 3m 1 m kip/ft
1
P1 P2
4 kip/ft
(½)(4 kip/ft)(9ft) = 18 kip at 3 ft
La Lb
A B
(1 kip/ft)(16ft) = 16 kip at 22 ft
9 ft 6 ft 10 ft A B
Ra Rb
5 ft 9 ft

Exercise 13:
 Determinethe resultant force of the loads acting
on the beam shown, and specify its location on
the beam with respect to point A.


 Determine the
resultant force of
the loads acting
on the beam
shown, and
specify its location
on the beam with
respect to point A.

Exercise:
 Considering that the 2 force systems shown are
equivalent (same mechanical effect), determine the
tension in each of the equal length cables (left diagram)
supporting a 1250 lb weight.

I P
S K
Equivalent
Force
Systems

Exercise: Solution
 Determine the tension in each of the equal length cables
supporting a 1250 lb weight.
For equilibrium, the resultant R from the tension (T1, T2) in the two cables must be
equal and opposite force vector from the 1250 lb weight.

IP
70° Solve for force triangle using
law of Sines:

S K 35° T2 T1 / (sin 35°) = R / (sin 110°)

110° T1 = R sin 35° / sin 110°
R
35° T1 T1 = (1250)(0.5736)/(0.9397)

T1 =762.98 lb
1250 lb 1250 lb
T1 =763 lb T2 =763 lb

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