Engineering Mechanics: STATICS

1
Engineering MechanicsSTATICS

• Anthony Bedford and Wallace Fowler
• SI Edition

Teaching Slides Chapter 4 System of Forces
Moments
2
Chapter Outline

• 2-Dimensional Description of the Moment
• The Moment Vector
• Moment of a Force About a Line
• Couples
• Equivalent Systems
• Representing Systems by Equivalent Systems
• Computational Mechanics

3
4.1 2-D Description of the Moment

• Consider a force of magnitude F a point P
  view them in a direction perpendicular to the
  plane containing the force vector the point
• The magnitude of the moment of the force about P
  is the product DF, where D is the perpendicular
  distance from P to the line of action of the
  force
• The force will tend to cause counterclockwise
  rotation about point P

4
4.1 2-D Description of the Moment

• Imagine that the force acts on an object that can
  rotate about point P, the force would cause
  counterclockwise rotation
• Direction of the moment is counterclockwise
• Convection
• Counterclockwise moments positive
• Clockwise moments negative
• Moment of force about P
• Mp DF
  (4.1)

5
4.1 2-D Description of the Moment

• If the line of action of F passes through P, the
  perpendicular distance D 0 the moment of F
  about P is zero
• The dimensions of the moment are
  (distance) ? (force)
• E.g. moments can be expressed in newton-meters in
  SI units

6
4.1 2-D Description of the Moment

• Intuitively, we know that the attachment of the
  shelf to the wall is more likely to fail if you
  place the television set away from the wall

7
4.1 2-D Description of the Moment

• The magnitude direction of the force exerted on
  the shelf by the weight of the television are the
  same in each case but the moments exerted on the
  attachment are different
• Moment exerted about P by its weight when it is
  near the wall, MP ?D1W, is smaller in magnitude
  than the moment about P when it is placed away
  from the wall, MP ?D2W

8
4.1 2-D Description of the Moment

• This method can be used to determine the sum of
  the moments of a system of forces about a point
  if the forces are 2-D (coplanar) the point lies
  in the same plane
• Sum of the moments exerted about point P by the
  load W1 the counterweight W2
• ? Mp D1W ? D2W

9
4.1 2-D Description of the Moment

• This moment tends to cause the top of the
  vertical tower to rotate could cause it to
  collapse
• If D2 is adjusted so that D1W D2W, the moment
  about point P due to the load the counterweight
  is zero
• If a force is expressed in terms of components,
  the moment of the force about a point P is equal
  to the sum of the moments of its components about
  P

10
Example 4.1 Determining the Moment of a Force
What is the moment of the 40-kN force in Fig. 4.4
about point A?
Strategy We can calculate the moment in 2 ways
by determining the perpendicular distance from
point A to the line of action of the force or by
expressing the force in terms of components
determining the sum of the moments of the
components about A.
11
Example 4.1 Determining the Moment of a Force
Solution 1st Method The perpendicular distance
from A to the line of action of the force is
D (6 m) sin 30 3 m
The magnitude of the moment of the force about A
is (3 m)(40 kN) 120 kN-m the direction of the
moment about A is counterclockwise. Therefore,
the moment is MA
120 kN-m
12
Example 4.1 Determining the Moment of a Force
Solution 2nd Method Express the force in terms
of horizontal vertical components The
perpendicular distance from A to the line of
action of the horizontal component is zero, so
the horizontal component exerts no moment about A.
13
Example 4.1 Determining the Moment of a Force
Solution The magnitude of the moment of the
vertical component about A is (6
m)(40 sin30 kN) 120 kN-m the direction of
the moment about A is counterclockwise. The
moment is MA 120 kN-m
14
Example 4.1 Determining the Moment of a Force

• Critical Thinking
• In the 1st method, we calculated the moment of
  the 40-kN force about A by determining the
  perpendicular distance to the line of action of
  the force multiplying it by the magnitude of
  the force
• In the 2nd method, we 1st expressed the 40-kN
  force in terms of components then calculated
  the sum of the moments of the components about A

15
Example 4.1 Determining the Moment of a Force

• Critical Thinking
• The 2 methods yield the same answer
• Demonstration of Varignons Theorem
• It is not always easy to determine the
  perpendicular distance to the line of action of
  the force
• In some cases, it will be much simpler to
  determine the moment of a force by calculating
  the sum of the moments of its components

16
Example 4.2 Moment of a System of Forces
4 forces act on the machine part in Fig 4.5. What
is the sum of the moments of the forces about the
origin O?
Strategy We can determine the moments of the
forces about O directly from the given
information except for the 4-kN force. We will
determine its moment by expressing it in terms of
components summing the moments of the
components.
17
Example 4.2 Moment of a System of Forces
Solution Moment of the 3-kN Force The line of
action of the 3-kN force passes through O. It
exerts no moment about O. Moment of the 5-kN
Force The line of action of the 5-kN force also
passes through O. It too exerts no moment about
O.
18
Example 4.2 Moment of a System of Forces
Solution Moment of the 2-kN Force The
perpendicular distance from O to the line of
action of the 2-kN force is 0.3 m the direction
of the moment about O is clockwise. The moment of
the 2-kN force about O is
?(0.3 m)(2 kN) ?0.600 kN-m (Notice that we
converted the perpendicular distance from
millimeters into meters, obtaining the result in
terms of kilonewton-meters)
19
Example 4.2 Moment of a System of Forces
Solution Moment of the 4-kN Force Introduce a
coordinate system express the 4-kN force in
terms of x y components
The perpendicular distance from O to the line of
action of the x component is 0.3 m the
direction of the moment about O is clockwise. The
moment of the x component about O is
?(0.3 m)(4 cos 30 kN) ?1.039 kN-m
20
Example 4.2 Moment of a System of Forces
Solution The perpendicular distance from O to the
line of action of the y component is 0.7 m the
direction of the moment about O is
counterclockwise. The moment of the y component
about O is (0.7 m)(4 sin 30 kN)
1.400 kN-m The sum of the moments of the 4
forces about point O is ? MO ?0.600 ?
1.309 1.400 ?0.239 kN-m The 4 forces exert a
0.239 kN-m clockwise moment about point O.
21
Example 4.2 Moment of a System of Forces

• Critical Thinking
• If an object is subjected to a system of known
  forces, it is useful to determine the sum of the
  moments of the forces about a given point
• Provides a test for equilibrium ? an object is in
  equilibrium only if the sum of the moments about
  any point is zero
• (Notice that the object in this example is not
  in equilibrium)
• Furthermore, in dynamics the sum of the moments
  of the forces acting on objects must be
  determined in order to analyze their angular
  motions

22
Example 4.3 Summing Moments to Determine an
Unknown Force
The weight W 300 N (Fig. 4.6). The sum of the
moments about C due to the weight W the force
exerted on the bar CA by the cable AB is zero.
What is the tension in the cable?
23
Example 4.3 Summing Moments to Determine an
Unknown Force
Strategy Let T be the tension in the cable AB.
Using the given dimensions, express the
horizontal vertical components of the force
exerted on the bar by the cable in terms of T.
Then by setting the sum of the moments about C
due to the weight of the bar the force exerted
by the cable equal to zero, we can obtain an
equation for T.
24
Example 4.3 Summing Moments to Determine an
Unknown Force
Strategy
Using similar triangles, express the force
exerted on the bar by the cable in terms of
horizontal vertical components. The sum of the
moments about C due to the weight of the bar
the force exerted by the cable AB is
Solving for T T 0.357W 107.1 N
25
Example 4.3 Summing Moments to Determine an
Unknown Force

• Critical Thinking
• This example is a preview of the applications we
  consider in Chapter 5 demonstrates why you must
  know how to calculate moments of forces
• If the bar is in equilibrium, the sum of moments
  about C is zero ? this condition allowed us to
  determine the tension in the cable
• We do not need to consider the force exerted on
  the bar by its support at C since the moment of
  that force about C is zero

26
4.2 The Moment Vector

• The moment of a force about a point is a vector
• Using the description in Section 4.1, we are
  specifying the magnitude direction of the
  moment vector
• Consider a force vector F a point P
• Moment of F about P is the vector
• MP r ? F
  (4.2)
• where r is a position vector from P to any point
• on the line of action of F

27
4.2 The Moment Vector

• Magnitude of the Moment
• From the definition of the cross product, the
  magnitude of MP is
• MP rF sin ?
• where ? is the angle between the vectors r
  F when
• they are placed tail to tail
• The perpendicular distance from P to the line of
  action of F is D r sin ?

28
4.2 The Moment Vector

• Magnitude of the moment MP equals the product of
  the perpendicular distance from P to the line of
  action of F the magnitude of F
• MP DF
  (4.3)
• Notice that if we know the vectors MP F, this
  equation can be solved for the perpendicular
  distance D

29
4.2 The Moment Vector

• Direction of the Moment
• From the definition of the cross product, MP is
  perpendicular to both r F
• MP is perpendicular to the plane containing P F
• Denoted by a circular arrow around the vector

30
4.2 The Moment Vector

• The direction of MP also indicates the direction
  of the moment
• Pointing the thumb of the right hand in the
  direction of MP, the arc of the fingers
  indicates the direction of the rotation that F
  tends to cause about P

31
4.2 The Moment Vector

• The result obtained from Eq. (4.2) doesnt depend
  on where the vector r intersects the line of
  action of F
• r r u
• r ? F (r u) ? F r ? F
• Because the cross product of the parallel
• vectors u F is zero

32
4.2 The Moment Vector

• In summary, the moment of a force F about a point
  P has 3 properties
• 1. The magnitude of MP the product of the
• magnitude of F the perpendicular distance
• from P to the line of action of F. If the
  line of
• action passes through P, MP 0.
• 2. MP is perpendicular to the plane containing P
  F.
• 3. The direction of MP indicates the direction of
  
• the moment through a right-hand rule. Since
• the cross product is not commutative, it is
• essential to maintain the correct sequence of
  
• the vectors in the equation MP r ? F

33
4.2 The Moment Vector

• To determine the moment of the force F about the
  point P
• The vector from P to the point of application of
  F
• r (12 ? 3)i (6 ? 4)j (?5
  ? 1)k
• 9i 2j ?6k (m)

34
4.2 The Moment Vector

• The moment is
• The magnitude of MP
• equals the product of the magnitude of F the
  perpendicular distance D from point P to the line
  of action of F

35
4.2 The Moment Vector

• Therefore,
• The direction of MP tells us both the orientation
  of the plane of the plane containing P F the
  direction of the moment

36
4.2 The Moment Vector

• Relation to the 2-D Description
• If our view is perpendicular to the plane
  containing the point P the force F, the 2-D
  description of the moment in Section 4.1
  specifies both the magnitude direction of the
  vector MP
• In this situation, MP is perpendicular to the
  page the right-hand rule indicates whether it
  points out or into the page

37
4.2 The Moment Vector

• E.g. the view is perpendicular to the x-y plane
  the 10-N force is contained in the x-y plane
• The perpendicular distance from O to the line of
  action of the force is 4 m
• The 2-D description of the moment of the force
  about O its magnitude is (4 m)(10 N) 40 N-m
  its direction is counterclockwise or MO 40 N-m

38
4.2 The Moment Vector

• Therefore, the magnitude of the vector MO is 40
  N-m the right-hand rule indicates that it
  points out of the page
• MO 40k (N-m)
• We can confirm this result by using Eq. (4.2).
  Let r be the vector from O to the point of
  application of the force
• MO r ? F (4i 2j) ? 10j 40k (N-m)
• As this example illustrates, the 2-D description
  of the moment determines the moment vector
• The converse is also true

39
4.2 The Moment Vector

• Varignons Theorem
• Let F1, F2, , FN be a concurrent system of
  forces whose lines of action intersect at a
  point Q
• The moment of the system about a point P is
• (rPQ ? F1) (rPQ ? F2) ??? (rPQ ?
  FN)
• rPQ ? (F1 F2 ???
  FN)
• where rPQ is the vector from P to Q

40
4.2 The Moment Vector

• This result, known as the Varignons Theorem,
  follows from the distributive property of the
  cross product, Eq. (2.31)
• It confirms that the moment of the force about a
  point P equal to the sum of the moments of it
  components about P

41
Example 4.4 2-D Description the Moment Vector

• Determine the moment of the 400-N force in
• Fig. 4.13 about O.

42
Example 4.4 2-D Description the Moment Vector

• Strategy
• We will determine the moment in 2 ways
• (a) Use the 2-D description of the moment.
  Express
• the force in terms of its components
  determine
• the moment of each component about O by
• multiplying the magnitude of the component
  
• the perpendicular distance from O to its
  line of
• action.
• (b) Obtain the vector description of the moment
  by
• using Eq. (4.2).

43
Example 4.4 2-D Description the Moment Vector

• Solution
• (a) Expressing the force in terms of horizontal
  
• vertical components, the 2-D description of
  the
• moment is
• MO ?(2 m)(400 cos 30 N) ? (5 m)(400 sin 30 N)
• ?1.69 kN-m

44
Example 4.4 2-D Description the Moment Vector

• Solution
• (b) To apply Eq. (4.2), introduce the coordinate
• system shown
• Choose the Vector r
• Let r be the vector from O to the point of
• application of the force
• r 5i 2j (m)

45
Example 4.4 2-D Description the Moment Vector

• Solution
• (b) Evaluate r ? F
• The moment is
• MO r ? F
• (5i 2j) ? (400 cos 30i ? 400
  sin 30j)
• ?1.69k (kN-m)

46
Example 4.4 2-D Description the Moment Vector

• Critical Thinking
• In most 2-D situations, it is easier to use the
  2-D description of the moment than the vector
  description
• However, studying the relationship between the
  2-D vector descriptions of the moment provides
  insight into the vector description
• Demonstrates, for this special case, that the
  magnitude direction of the vector specify the
  magnitude direction of the moment
• In 3-D situations, the vector description of the
  moment is nearly always used

47
Example 4.5 Determining the Moment the
Perpendicular Distance to the Line of Action

• The line of action of the 90-N force F in Fig.
  4.14
• passes through points B C.
• (a) What is the moment of F about point A?
• (b) What is the perpendicular distance from point
  A
• to the line of action of F?

48
Example 4.5 Determining the Moment the
Perpendicular Distance to the Line of Action

• Strategy
• We must use Eq. (4.2) to determine the moment.
  
• Since r is a vector from A to any point on
  the line
• of action of F, we can use either the vector
  from A
• to B or the vector from A to C. To
  demonstrate that
• we obtain the same results, we will
  determine the
• moment using both.
• Since the magnitude of the moment is equal to
  the
• product of the magnitude of F the
  perpendicular
• distance from A to the line of action of F,
  we can
• use the result of (a) to determine the
  perpendicular
• distance.

49
Example 4.5 Determining the Moment the
Perpendicular Distance to the Line of Action

• Solution
• To evaluate the cross product in Eq. (4.2), we
  need the components of F.
• The vector from B to C is
• (7 ? 11)i (7 ? 0)j (0 ? 4)k ?4i
  7j ? 4k (m)

Dividing this vector by its magnitude ? unit
vector eBC that has the same direction as F
50
Example 4.5 Determining the Moment the
Perpendicular Distance to the Line of Action

• Solution
• Now express F as the product of its magnitude
  eBC
• F 90eBC ?40i 70j ? 40k
  (N)
• Choose the vector r
• The position vector from A to B is
• rAB (11 ? 0)i (0 ? 6)j (4 ? 5)k
• 11i ? 6j ? k (m)

51
Example 4.5 Determining the Moment the
Perpendicular Distance to the Line of Action

• Solution
• (a) Evaluate r ? F
• The moment of F about A is

52
Example 4.5 Determining the Moment the
Perpendicular Distance to the Line of Action

• Solution
• (a) Alternative Choice of Position Vector
• If we use the vector from A to C instead
• rAC (7 ? 0)i (7 ? 6)j (0 ? 5)k
• 7i 7j ? 5k (m)
• We obtain the same result

53
Example 4.5 Determining the Moment the
Perpendicular Distance to the Line of Action

• Solution
• (b) The perpendicular distance is

54
Example 4.5 Determining the Moment the
Perpendicular Distance to the Line of Action

• Critical Thinking
• Suppose that you were unfamiliar with the vector
  description of the moment
• The magnitude of the moment of the force F about
  the point A equals the product of the
  perpendicular distance from A to the line of
  action of F with the magnitude of F
• To determine the perpendicular distance ? vector
  description of the moment
• The magnitude of the moment vector is the product
  of the perpendicular distance from A to the line
  of action of F with the magnitude of F
  direction of the moment vector defines the
  direction of the moment

55
Example 4.6 Applying the Moment Vector

• The cables AB AC in Fig. 4.15 exerted from
  an attachment point A on the floor to attachment
  points B C in the walls. The tension in cable
  AB is 10 kN the tension in cable AC is 20 kN.
  What is the sum of the moments about O due to the
  forces exerted on the attachment point A by the
  2 cables?

56
Example 4.6 Applying the Moment Vector

• Strategy
• We must express the forces exerted on the
  attachment point A by the 2 cables in terms of
  their components.
• Then we can use Eq. (4.2) to determine the
  moments the forces exert about O.

57
Example 4.6 Applying the Moment Vector

• Solution
• Let FAB FAC be the forces exerted on the
  attachment point A by the 2 cables.
• To express FAB in terms of its components, we
  determine the position vector from A to B
• (0 ? 4)i (4 ? 0)j (8 ? 6)k
• ? 4i 4j 2k (m)
• Divide it by its magnitude ?
• unit vector eAB with the same
• direction as FAB

58
Example 4.6 Applying the Moment Vector

• Solution
• Express the force FAC in terms of its
  components in the same way

59
Example 4.6 Applying the Moment Vector

• Solution
• Choose the Vector r
• Since the lines of action of both forces
  passes through point A, we can use the vector
  from O to A to determine the moments of both
  forces about point O r 4i 6k (m)
• Evaluate r ? F
• The sum of moments is

60
Example 4.6 Applying the Moment Vector

• Critical Thinking
• The lines of action of the forces FAB FAC
  intersect at A
• Notice that, according to Varignons theorem, we
  could summed the forces 1st, obtaining
• FAB FAC ?0.952i 15.24j ? 13.81k (kN)

61
Example 4.6 Applying the Moment Vector

• Critical Thinking
• And then determined the sum of the moments of the
  2 forces about O by calculating the moment of the
  sum of the 2 forces about O