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MOMENT OF A FORCE ABOUT A POINT

• The tendency of a body to rotate about an axis

passing through a specific point O when acted

upon by a force (sometimes called a torque).

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APPLICATIONS

A torque or moment of 12 N · m is required to rotate the wheel.

Which one of the two grips of the wheel above will require less

force to rotate the wheel?

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Couple Moment

• The couple-moment is known as a free vector, meaning that it can be moved

anywhere in space without changing its

meaning.

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100mm

100mm

150mm

100mm

135 N

135 N

90 N

90 N

100mm

135 N

135 N

3.14 Equivalent Couples

Figure shows three couples which act successively on the same rectangular box. As seen in the

preceding section, the only motion a couple can impart to a rigid body is a rotation.

Since each of the three couples shown has the same moment M ( same direction and same magnitude

M = 135 N. m ), we can expect the three couples to have the same effect on the box.

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MOMENT OF A FORCE ABOUT A POINT

• Magnitude of a moment

Mo = F d N.m

• Mo = Magnitude of the moment of F around point O

• d = Perpendicular distance from O to the line of action

of F

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• Moment produces a rotation.

• Direction determined by using the Right-

Hand Rule.

• The thumb points along the moment axis

and the other fingers are curled following

the sense of rotation.

• Could be ‘clockwise’ (CW) or ‘anti/counter

clockwise’ (CCW).

DIRECTION OF MOMENT OF

A FORCE

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Direction of Moment

Choose the convenient sense of rotation for each analysis.

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Calculating moment

Scalar Analysis

Mo = F d

Mo = F (r sin θ)

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Sample problem 3.1

A 450N vertical force is applied to the end of a lever

which is attached to a shaft at O.

Determine

(a) the moment of the 450N force about O

(b) 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 the shaft a 1100N vertical force must act

to create the same moment about O

(e) whether any one of the forces obtained in parts b,c and

d is equivalent to the original force.

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Solution

a) Moment about O

The perpendicular distance from O to the line

of action of the 450N force is

The magnitude of the moment about O of the

450N force is

d = (0.6m) cos 60˚ = 0.3 m

Mo = Fd = (450 N) (0.3m) = 135 N.m

Since the force tends to rotate the lever clockwise

about O, the moment will be represented by a

vector Mo perpendicular to the plane of the figure and

pointing into the paper.We express this fact by writing.

Mo = 135 N.m

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b. Horizontal Force

In this case, we have

d = (0.6m) sin 60˚ = 0.52 m

Since the moment about O must be 135N.m,

we write

Mo = Fd

135 N.m = F (0.52 m)

F = 260 N F= 260 N

Solution

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c. Smallest Force

Since Mo = Fd, the smallest value of F occurs

when d is maximum. We choose the force perpendicular

to OA and note that d = 0.6m, thus

Mo = Fd

135 N.m = F (0.6 m)

F = 225 N F= 225 N 30˚

Solution

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d. 1100 N Vertical Force

In this case Mo = Fd yields

135 N.m = (1100 N)d

d = 0.12 m

OB cos 60˚ = d

OB = d

cos 60˚

OB = 0.24 m

Solution

e. None of the forces considered in parts b,c and d is equivalent to

the original 450N force. Although they have the same moment about O,

they have different x and y components.

In other words, although each forces tends to rotate the shaft in the same

manner, each causes the lever to pull on the shaft in a different way.

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Sample problem 3.2

A force of 800 N acts on as bracket as shown.

Determine the moment of a force about B.

16 0.2

0.16

Fx= 800 cos 60

Fy = 800 sin 60

B

A

MB = xFy + yFx

= 0.2 (800 sin 60)

+ 0.16 (800 cos 60)

= 138.564 + 64

= 202.6 Nm

Solution

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Sample problems 3.3

A 135N force acts on the end of the 0.9 m

lever as shown.

Determine the moment of the force about O 135N

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135N

Solution

The force is replaced by two components, one component

P in the direction of OA and one component Q perpendicular

to OA.

Since O is on the line of action of P, the moment of P about O

is zero and the moment of the 135N force reduces to the moment

of Q, which is clockwise and, thus, is represented by a negative

scalar.

Q = (135N) sin 20˚ = 46.2 N

Mo = -Q (0.9m) = -(46.2N)(0.9m) = -41.6N.m

Since the value obtained for the scalar Mo is negative,

the moment Mo points into the paper. We write

Mo = 41.6 N.m

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EXAMPLE 1

Given: A 400 N force is

applied to the frame

and = 20°.

Find: The moment of the

force at A.

1) Resolve the force along x and y axes.

2) Determine MA using scalar analysis.

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+ Fy = -400 sin 20° N

+ Fx = -400 cos 20° N

+ MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m

= 1160 N·m

Solution

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Problems 3.1

A 90-N Force is applied to the control rod AB as shown. Knowing that the length of the rod is 225mm,

determine the moment of the force about point B by resolving the force into components along AB and in a

direction perpendicular to AB.

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Solution

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Problems 3.2

A 90-N Force is applied to the control rod AB as shown. Knowing that the length of the rod is 225mm,

determine the moment of the force about point B by resolving the force into horizontal and

vertical components.

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Solution 3.3

3.4 m

4.8 m

A 3N force P is applied to the lever which controls the auger of a snow blower. Determine the moment of P about

A when αis equal to 30º.

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Equivalent System

MOVING A FORCE ITS LINE OF ACTION ON

Sliding VECTOR

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MOVING A FORCE OF ITS LINE OF ACTION OFF

Free VECTOR

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Resultant of a Force and Couple System

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3.20 Further Reduction of a Force and Couple System

If R and MRP are perpendicular to each other,

the force-couple system at P can be further reduced to a

single resultant force.

Will be the case for system consisting either :

(a) concurrent force

(b) coplanar force

(c) parallel forces

(a) concurrent force

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(b) Coplanar force systems

(c) Parallel force systems

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The moment of the resultant force about the grip

the moment of all the forces about the grip

FRd = F1d1 + F2d2 + F3d3

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Sample problems 3.8

A 4.80m long beam is subjected to the forces shown.

Reduce the given system of forces to

a) an equivalent force couple system at A

b) single force or resultant

150N 600N 100N 250N

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use FRd = F1d1 + F2d2 + F3d3 ;

solution

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= - 600(1.6m) + 100(2.8m) – 250(4.8)

= - 1880 Nm

+ FR = 150 - 600 + 100 - 250 = - 600 N

= 600 N

+ SMA = F1d1 + F2d2 + F3d3

F1 F2 F3

solution 150N 600N 100N 250N

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A B

d

FR = 600 [N]

FRd = SMA

(600) d = 1880 Nm

d = 1880/600

= 3.13 m

Thus, the single force equivalent to the given system;

FR = 600 [N] , d = 3.13 [m]

solution

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Solution

16.4 N 16.4 N 14 N

D

A B C =

D

L R

2.1 m

0.85 m d 0.25 m

For equivalence,

F : - 16.4 – 16.4 – 14 = - R

R = 46.8 N

: - (0.25)(16.4) – (1.10)(16.4) – (1.10 + d)(14) = - L (46.8)

37.54 + 14d = 46.8 L »Eq 1

S

DM

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Example 3 (cont)

(a)For d = 0.625 m, substitute the value of d in Eq 1,

37.54 + 14 (0.625) = 46.8 L

L = 0.9891 m

The resultant passes through a point 0.9891 m to the right

of D

(b) For L = 1.05 m, substitute the value of L in Eq. 1,

37.54 + 14d = 46.8 (1.05)

d = 0.8275 m

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THE END