This is the force system in which lines of action of individual forces lie in the same plane but act at different point of application.

CHAPTER – II RESULTANT OF COPLANAR NON CONCURRENT

FORCE SYSTEM

Fig. 2

F1 F2

F5F4

F3

Fig. 1

F2F1

F3

Coplanar Non-concurrent Force System:

TYPES

1. Parallel Force System – Lines of action of individual forces are parallel to each other.

Fig. 1

F2F1

F3

2. Non-Parallel Force System – Lines of action of the forces are not parallel to each other.

Fig. 2

F1 F2

F5F4

F3

MOMENT OF A FORCE ABOUT AN AXIS

The applied force can also tend to rotate the body aboutan axis in addition to motion. This rotational tendency is known as moment.

This is a vector quantity having both magnitude and direction.

Definition: Moment is the tendency of a force to make a

rigid body to rotate about an axis.

Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising moment arm and line of action of the force.

BAd

F

Distance AB = d.

Moment Arm: Perpendicular distance from the line of action of the force to moment center.

Moment Center: This is the position of axis on co-planar system. (B).

Consider the example of pipe wrench.

The force applied which is

perpendicular to the handle of the

wrench tends to rotate the pipe about

its vertical axis. The magnitude of

this tendency depends both on the

magnitude of the force and the

effective length ‘d’ of the wrench

handle. Force perpendicular to the handle is more effective.

EXAMPLE FOR MOMENT

MAGNITUDE OF MOMENT

B

F

Ad

It is computed as the product of the of the force and the perpendiculardistance from the line of action to the point about which moment iscomputed. (Moment center).

Unit – Unit of Force × Unit of distance kN-m, N-mm etc.

MA = F×d

SENSE OF MOMENT

‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the sense of moment vector’.

M M

For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +ve for counterclockwise and –ve for clockwise rotations or vice-versa.

The sense is obtained by ‘Right Hand Thumb’ rule.

VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)

Statement: The moment of a force about a moment center or axis is equal to the algebraic sum of the moments of component forces about the same moment center (axis).

‘P’ & ‘Q’ are component forces of ‘R’.

Ry

Py

Q R

A O

p

q r PQy

Y

X

Proof (by Scalar Formulation):Proof (by Scalar Formulation):

Let ‘R’ be the given force.

, and are the inclinations of‘P’, ‘R’ and ‘Q’ respectively w.r.t X – axis.

p, r and q are moment arms from ‘O’ of P, R and Q respectively.

‘O’ is the moment center.

We have, Ry = Py + Qy

Moment of resultant R about O = algebraic sum of moments of component forces P & Q about same moment center ‘O’.

i.e., R × r = P × p + Q × q

From (1), R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)

From le AOD, q/AO = Sin From le AOC, r/AO = Sin From le AOB, p/AO = Sin

R Sin = P Sin + Q Sin ----(1)

Ry

Py

Q R

A O

p

q r PQy

Y

X

VARIGNON’S THEOREM – PROOF BY VECTOR FORMULATION

Consider three forces F1, F2, and F3 concurrent at point ‘A’ as shown in fig. Let r is the position vector from ‘O’ to point ‘A’.

where, R is the resultant of three original forces.

Mo = r × (F1+F2+F3) = r × R

By the property of cross product,

Mo = ∑(r×F) = (r×F1) + (r×F2) + (r×F3).

The sum of moments about ‘O’ for these three forces by cross-product is,

1. Simplifies the computation of moments by judiciously selecting the moment center.

Moment can be determined by resolving a force into x & y components because finding x & y distances in many circumstances may be easier than finding perpendicular distance from moment center to line of action (d).

APPLICATIONS OF VARIGNON’S THEOREM

Location of resultant - location of line of action of resultant in case of non-concurrent force system which is an additional information required in the resultant problems.

2.

COUPLE

Two parallel, non collinear (separated by certain

distance) forces that are equal in magnitude and opposite

in direction form ‘couple’.

d

F

FHence, couple does not produce any translation and

produces only rotation.

The algebraic summation of the

two forces forming couple is zero.

Moment of a Couple: Consider two equal and opposite forces separated by a distance ‘d’.

O

da

F

F

The moment of a couple about any point is constant and is equal to the product of one of the forces and the perpendicular distance between them.

Thus, the moment of the couple about ‘O’ is independent of the location, as it is independent of ‘a’.

+ ∑ Mo = F × ( a + d) – F × a = F× d

The sum of moments of two forces about the point ‘O’ is,

Let ‘O’ be the moment center at a distance ‘a’ from one of the forces.

RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM

P

F

Fig. (a)

F

=Q

M=F × d

Fig. (c)

=P

FF

Qd

Fig. (b)

F

F

Two equal and opposite forces of same magnitude F and

parallel to the force F at P are introduced at Q.

A given force F applied at a point can be replaced by an equal force applied at another point Q together with a couple which is equivalent to the original system.

P

F

=

P

F

Q

F

= Q

M=F × d

d

Fig. (a) Fig. (b) Fig. (c)

F

F

Of these three forces, two forces i.e., one at P and the

other oppositely directed at Q form a couple.

The system in Fig. ( c ) is equivalent to the system in Fig. ( a ).

Third force at Q is acting in the same direction as that at P.

Moment of this couple, M = F × d.

Thus, the force F acting at a point such as P in a rigid

body can be moved to any other given point Q, by adding a

couple M. The moment of the couple is equal to moment of

the force in its original position about Q.

60º

O

A

240

mm

F

A 100N vertical force is applied to the end of a lever

at ‘A’, which is attached to the shaft at ‘O’ as shown

in the figure.

PROBLEM - 1

Determine,

a) The moment of 100N force about ‘O’.

b) Magnitude of the horizontal force applied at ‘A’,

which develops same moment about ‘O’.

c) The smallest force at ‘A’, which develops same

effect about ‘O’.

d) How far from the shaft a 240N vertical force must

act to develop the same effect?

e) Whether any of the above systems is equivalent to

the original?

Perpendicular distance from the line of action of force F to moment center ‘O’ = d

SOLUTION

60º

O

A

240

mm

d

F

= 12,000 N-mm (Clockwise)

Moment about ‘O’ = F × d = 100 × 120

d = 240 cos 60º

=120 mm.

Case a) The moment of 100N force about ‘O’.

Perpendicular distance between the line of action of horizontal force F at A to moment center ‘O’,

OA

60º

240m

m

d

F

Therefore, F = 12,000/207.85 = 57.73 N

Moment about ‘O’ = F × d

= F × 207.85

= 12,000 N-mm (Clockwise)

d = 240 sin 60º = 207.85 mm.

Case b) Magnitude of the horizontal force applied at ‘A’, which develops same moment about ‘O’.

Case c) The smallest force at ‘A’, which develops same effect about ‘O’.

O

A

60º

d =

240

mm

F

Therefore, Fmin = 12,000/240 = 40N.

i.e., d = 240 mm.

Force is smallest when the perpendicular distance is maximum so as to produce same M.

F = M/d

Distance along x-axis, X = M/F = 12,000/240

= 50 mm. O

A

60º

X F

d

d = (X / cos 60) = 50/cos 60 = 100 mm

Case d) How far from the shaft a 240N vertical force

must act to develop the same effect?

Distance along the shaft axis

Case e) Observations:

II) Shaft rotates in the same manner but the pulling effect on

the shaft is different in different cases.

I) None of the above force system is equivalent to the

original even though all of them produce same moment.

PROBLEM - 2

A 50kN force acts on the corner of a 4m x 3m box as

shown in the Fig. Compute the moment of this force about

A by a) Definition of Moment

b) Resolving the force into components along CA

and CB.

A B

CD60º

3m

4m

F=50 kN

SOLUTION

a) By Definition of Moment:

A B

CD60º

36.87º60º

dE

3m

4m

F=50 kN

Moment about A = 50 × 1.96 = 98.20 kNm.

= 5 × sin 23.13º = 1.96 m.

From ∆le ACE, d = AC × sin ( ACE)

ACE = 60º – 36.87º = 23.13º

ECD = 60º

CAD = tan-1(3/4) = 36.87º

m534 22 AC =

To determine ‘d’:

b) By Components:

A B

CD60º

3m

4m

F=50kN

Fx

Fy

= + 98.20kNm.

= - 25 × 3 + 43.3 × 4

+ ΣMA = - Fx × 3 + Fy × 4

Fy = 50 × sin 60 = 43.30kN.

Fx = 50 × cos 60 = 25kN.

An equilateral triangle of sides 200mm is acted upon by 4 forces as shown in the figure. Determine magnitude and direction of the resultant and its position from point ‘D’.

PROBLEM - 3

30º

80kN

30kN

50kN

60kN

60º

200mmD

B

A

Resultant & its inclination:

SOLUTION

= tan-1(6.69/56.96) = 6.7º = tan-1(Ry/Rx)

Inclination w.r.t horizontal = θR

R =

+ ΣFy = Ry

= +30 + 60 cos30º – 50 cos60º + 0.

+ ΣFx = Rx

= +56.96kN

Resolving forces we have,

= -80 + 60 sin30º + 50 sin60º + 0 = -6.69kN.

30º

80kN

30kN

50kN

60kN

60º

60 cos30

60 sin30

50 sin 60

50 cos 60

200mm

A

B

D

b) Position w.r.to D:

d = 2000/57.35 = 34.87mm.

= 57.35 × d.

where ‘d’ = perpendicular distance from point D to the line of action of R.

= R × d

+ MD = -50 sin60 × 100 + 50 cos60 × (sin60 × 200)

- 60 × 100 + 80 × 100 + 30 × 0 = 2000kNmm.

Moment of the component forces about D:30º

80kN

30kN

50kN

60kN

60º

60 cos30

60 sin30

50 sin 60

50 cos 60

200mm

A

B

D

Location of the resultant along x-axis from the point D

30º

80kN

30kN

50kN

60kN

60º

200mmD

B

A

Rx

ΣFx

ΣFy

M= (ΣFy × x) x=M/ ΣFy

M= (ΣFy × x – ΣFx × y)--------- (1)If we resolve the resultant at a point on the x-axis, then the moment of force ΣFx about the point D, becomes zero, and

M= R × d

Location of the resultant along x-axis from the point D

30º

80kN

30kN

50kN

60kN

60º

200mmD

B

A

R

ΣFx

ΣFy Y

Location of the resultant along x-axis from the point D

M= - (ΣFx × y)

Location of the resultant along y-axis from the point D

M= (ΣFy × x)

Fy

Mx

Fx

My

PROBLEM - 4

Find the resultant and its position w.r.t ‘O’ of the non-concurrent system of forces shown in the figure.

F3=1000N

F2=500N

F1=2500N

F5=2000N

F4=1500N1m

1m

Ө2Ө4

1

1

Ө5

O

SOLUTION

A) To find the resultant –

F3=1000N

F2=500NF1=2500N

F5=2000N

F4=1500N1m

1m

Ө2Ө4

1

1

Ө5

O

= 2500+500 sin26.56-1500 sin56.31+2000 sin45 =2889.70N ↑

+↑ΣFy = Ry= F1+F2 sin Ө2-F4 sin Ө4+F5 sin Ө5

= -799.03N = 799.03N←= 500 × cos26.56 + 1000 –1500 × cos56.31-2000 × cos45

+ ΣFx = Rx = F2 cosӨ2 +F3 -F4 cosӨ4-F5 cos Ө5

Ө2 = tan-1(1/2) = 26.56°

Ө4 = tan-1(3/2) = 56.31°

Ө5 = tan-1(1/1) = 45°

∴ Resultant R =

x

y

R

R

Rx

RyR

ӨR

d = 1.43 m from O.

+ Mo =R×d = +2500×2 + 500×sin26.56×5 – 500× cos26.56×3 - 1000×1+1500× cos56.31×0 –1500×sin56.3×1+2000× cos45×1-2000×sin45×0

= 2998.14 × d

By Varignon’s theorem, Moment of the resultant about ‘O’ is = algebraic sum of the moments of its components about ‘O’.

B) Position of Resultant w.r.t ‘O’:

ӨR = tan-1 = tan-1(2889.7/799.03) = 74.54°

= 2998.14N

Determine the resultant of three forces acting on a dam section shown in the figure and locate its intersection with the base. Check whether the resultant passes through the middle one-third of the base.

PROBLEM - 5

60º

30 kN

120 kN

50 kN

6 m

1 m2 m

3 m

A B

SOLUTION

+ ∑Fx = Rx = 50 – 30 × cos 60 = 35 kN = 35 kN

6 m

60º

30 kN

120 kN

50 kN

1 m

2 m3 m 60º

A B

kNRR yx 12.15098.14535 2222 Resultant, R=

+ ∑Fy = Ry = -120 – 30 × sin 60 = -145.98 kN

= 145.98 kN

θR= tan-1(Ry/Rx) = tan-1(145.98/35) = 76.52º

Hence, the resultant passes through the middle 1/3rd of the base.

2m<3.53<4m

Middle 1/3rd distance is between 2m and 4m.

From A, X = 6 – 2.46 = 3.53 m.

Therefore, X = 360/145.98 = 2.47m from B.

MB= 30×1 + 120 × (6-2) - 50 × 3 = Ry × X

360 = 145.98 × X

Location of the resultant w.r.t. B:

TYPES OF LOADS ON BEAMS

W kN/m

L

W kN/m

L

1. Concentrated Loads – This is the load acting for very small length of the beam.2. Uniformly distributed load – This is the load acting for a considerable length of the beam with same intensity of W kN/m throughout its spread.Total intensity = W × L (acts at L/2 from one end of the spread)3. Uniformly varying load – This load acts for a considerable length of the beam with intensity varying linearly from ‘0’ at one end to W kN/m to the other representing a triangular distribution.

Total intensity of load = area of triangular = 1/2× W × L. (acts at 2×L/3 from ‘Zero’ load end)

 Locate the resultant w. r. to point A.

PROBLEM - 6

 SOLUTION:

Rx

Ry

R

θR

2m 1m 2m

A B30kN

40kN-m20kN/m

3m

45°

θR=tan-1(Ry/Rx) = tan-1(61.21/21.21) = 70.89º

R =

+ ΣFy = Ry = -30 sin45 – 20x2 = -61.21kN = 61.21kN↓

+ ΣFx = Rx = -30 cos45 = - 21.21kN = 21.21kN←

Position w.r.to A:

= 426.07/61.21 = 6.96m

X = M/Ry

The X-distance from A along the beam,

d = M/R = 426.0/64.78 = 6.58m

The perpendicular distance to the line of action of R from A

For this clockwise moment, the line of action must be onto the right of A.

+ MA = -40 – 30×cos45×0 – 30×sin45×(3+2)

– 20×2× (3+2+1+2/2) = -426.07 = 426.07 kNm

 Locate the resultant w. r. t point A.

PROBLEM - 7

 SOLUTION:

A B

2m 2m 1m 2m

1m

1m10kN

2kN/m

10kN

30° 5kN/m

kNRR yx 35.282766.8 2222 R =

+ ΣFy = Ry = -1/2×2×2-10-10×sin30-5×2 = -27kN =27kN↓

+ ΣFx = Rx = -10 cos30 = - 8.66 kN = 8.66kN←

θR=tan-1(Ry/Rx) = tan-1(27/8.66) = 72.22º Rx

R

θR

Ry

X= M/Ry = 124.06/27 = 4.59m

The X-distance from A along the beam,

The perpendicular distance to the line of action of R

from A, d = M/R = 124.06/28.35 = 4.38m.

For this clockwise moment, the line of action must be onto the right of A.

+ MA = - (1/2×2×2)×2/3×2 – 10×(2+2) + 10× cos30×1- 10 sin30×(2+2+1+1) – 5×2×(2+2+1+1/2×2) = -124.006kNm

= 124.006kNm

Position w.r.t A:

PROBLEM 8

A 50 N force is applied to the corner of a plate as shown in the fig. Determine an equivalent force - couple system at A. Also determine an equivalent force system consisting of a 150 N force at B and another force at A.

100 mm

50 mm

30 mm

30º 50N

B

A

SOLUTION

A) Force – Couple System at A:

100 mm

50 mm

30 mm

30º

Fy=50 cos 30

50N

Fx=50 sin 30

B

A

100 mm

50 mmA

Fy=50 cos 30

Fx=50 sin 30

Ma=3080N-mm

= -3080 N-mm.

= 3080 N-mm

= 25×50 - 43.3×100

+ ∑MA= Fx×50-Fy×100

Moment of the couple about A

These forces can be moved to A by adding the couple.

Fy = 50 × sin60= 43.3 N

Fx = 50 ×cos 60= 25 N.

SOLUTION

100 mm

50 mm

30 mm

30º

Fy=50 cos 30

50N

Fx=50 sin 30

B

A

100 mm

A

Fy=50 sin 60

Fx=50 cos60Ma=3080N-mm

SOLUTION

B) Forces at A and B :

∑ FAX = 50×cos60 - 150×cos46.8

=> FAX = -77.68N = 77.68N ←

The force at A is calculated as follows:

Therefore, θ = 46.8º.

i.e., MA = 150 × cosθ × 30 = 3080

The couple MA is because of two equal and opposite forces at A and B.

∑ FAY = -50×sin60 - 150×sin46.8

=> FAY = -152.65N = 152.65N

50 mm

100 mm

30 mmB

A

θ

150 N

FA

SOLUTION

FA = 171.28N

θA=tan-1(Fy/Fx) = tan-1(152.65/77.68) = 63.03º

100 mm

30 mmB

A

θ

150 N

171.28N

63.03

100 mm

50 mm

30 mm

30º 50N

B

A

=

PROBLEMS FOR PRACTICE

1. Determine the resultant of the parallel coplanar force system shown in fig.

(Ans. R=800N towards left, d=627.5mm)

400 N

1000 N

2000 N

600 N o

60º

60º 30º

10º

2. Four forces of magnitudes 10N, 20N, 30N and 40N acting respectively along the four sides of a square ABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A.

(Ans:R=28.28N, θ=45º, x=1.77a)

10N

40N

a

a

30N

20N

A B

CD

3. Four parallel forces of magnitudes 100N, 150N, 25N and 200N acting at left end, 0.9m, 2.1m and 2.85m respectively from the left end of a horizontal bar of 2.85m. Determine the magnitude of resultant and also the distance of the resultant from the left end.

(Ans:R=125N, x=3.06m)

4. Reduce the given forces into a single force and a couple at A. (Ans:F=320kN, θ=14.48º, M=284.8kNm)

100N80N

1m

1.5m

70.7kN200kN

A

30º45º

30º

5. Determine the resultant w.r.t. point A. (Ans:R=450kN, X=7.5kNm)

A

150N

1.5m3m1.5m

150Nm

500N100N