Forces are analyzed in a number of ways;

it is common approach to establish a coordinate system to quantify the forces and their effects in a system or body. Since it is customary to assign the axes, the analysis may be coplanar (two-dimensional) or non-coplanar (three-dimensional).

A system of forces may be represented by a resultant force which has the same effect as the system.

The resultant force, much like any other force, has magnitude and direction. The geometric sum of the forces will yield the resultant.

Forces on an object

Equivalent Resultant Force

2N

1N

3N

2N 4N

6N

Resultant =0

3N

6N

3N

Coplanar Force Systems analyze forces acting

on a body by taking their components along two designated axes.

A force system can be identified into two main types:

concurrent

non-concurrent.

Concurrent Forces are forces whose lines of action intersect at a common point. The resultant of concurrent forces originates from the intersection.

Point of Intersection

The resultant of concurrent forces must be defined by magnitude and direction. Magnitude represents the length of the vector while the direction is referred from the defined axis.

Resultant Force

Example 1: Compute the value of the resultant of the

concurrent system of forces shown.

300lb

100lb

200lb 400lb

60○

60○

1 2

45○

Example 2: In the concurrent force system shown in the figure.

Determine the value of P and the resultant force R acting at 30o to the left from the negative y-axis.

Example 3: The triangular block shown in the figure is subjected

to the loads P= 1600lbs and F = 600lbs. If AB= 8in and BC= 6in, resolve each load into normal and tangential components to AC.

A

P=1600lb

C θ B

F=600lb

Example 4: The block shown in the figure is acted upon by

its weight W=200lb, a horizontal force Q=600lb, and the pressure P exerted by the inclined plane. The resultant R of these forces is up and

parallel to the incline thereby

sliding the block up it.

Determine P and R.

(Hint: Take one axis

parallel to the incline)

Q

W

P

30○ 18○

Example 5: If the resultant force

is required to act along

the positive u-axis and

have a magnitude of

5 KN, determine the

required magnitude

of FB and its

direction θ.

Example 6: If the tension in rope AB is 100N, what is the

tension in rope BC? (Hint: The resultant of forces AB and BC is in the direction of the boat.)

Example 7: Two horses on opposite banks of canal pull a barge

moving parallel to the banks by means of two ropes. The tension in these ropes are 200lb and 240lb while the angle between them is 60 degrees. Find the

resultant pull on the

barge and the angle

between each of the

ropes and the sides of

the canal.

α

θ

Non-concurrent Forces are forces whose lines of action are parallel. The resultant of parallel force has a magnitude equal to the algebraic sum of the forces and is located somewhere between them.

5KN

10KN/m

25KN/m

10KN A B

3m 3m 1m

Resultant Force

The equivalent load of load diagrams – rectangular or uniform, triangular or uniformly varying – is equal to its area which is located at its geometric centroid.

To determine the location of the resultant, apply

Varignon’s Theorem – which states that the effect of the whole is equal to the sum of the effects of it components.

Using Varignon’s Theorem on taking moment about point A, the location can be found. Recall that the moment of a force about a point or axis is simply the magnitude of the force multiplied by its level arm or perpendicular distance to the point.

Also, remember that,

Moment = O if:

1. The Force intersects

the axis

2. The force is parallel to the axis

Example 8: Assuming clockwise moments to be positive,

compute for the moment of force F=450 lb and of force P=361 lb about points A,B,C, and D.

A

C

D B

F

P D

Example 9: Compute the resultant of the three forces shown in

the figure. Locate its intersection with the X and Y axes. (Hint: take also the moment of three forces in point O)

300 lb

390 lb

722 lb

O

y-axis

x-axis

5

12 30○

Example 10: Find the resultant of the force system and its

location from point A.

Example 11: Find the resultant of the non-concurrent force

system, its direction and location from point A.

5KN

10KN 10KN/m

10KN

A B

2m 1m 1m 3m

θ=60

Example 12: Find the value of P and F so that the four forces

shown in the figure produce an upward resultant force of 300 lb acting at 4 ft from the left end of the bar.

100lb 200lb P F

2 ft 2 ft 3 ft

Example 13: The resultant of three parallel loads (one load is

missing) is 30 lb acting up at 10 ft to the right of A. Compute the magnitude and position of the missing load.

40lb 60 lb

2 ft 11 ft A

Example 14: The 16-ft wing of an airplane is subjected to a lift

which varies from zero at the tip to 360lb/ft at the fuselage according to y=90x1/2, where x is measured from the tip. Compute the resultant and its location from the wing tip.

Example 15: Determine the resultant of three forces acting on the

dam shown and

locate its intersection

with the base AB. For

good design, this

intersection should

occur within the

middle third of

the base. Does it?

24000 lbs

Example 16: Serious neck injuries can occur when a football player

is struck in the face guard of his helmet in the manner shown, giving rise to a guillotine mechanism. Determine the moment

of the knee force P=50lb

about a point A. What

would be the magnitude

of the neck force F so

that it gives the counter-

balancing moment

about point A.

Example 17: Find the resultant of the non-concurrent force

system, its direction and location from point A.

A B

50KN

80KN 1m 2m 1.5m 1.5m 1m

10KN/m 10KN/m 10KN/m

20KN/m

Example 18: Find the resultant of the non-concurrent force

system, its direction and location from point A.

A

6 ft 4ft 40lb 50lb 3 ft

30 lb/ft 30 lb/ft

B

15 lb/ft

A couple is consists of two parallel, non-collinear forces that are equal in magnitude but oppositely directed.

Moment of a Couple = F*d = C

The moment of a couple is constant and independent of the moment center

F

F

d

X

O

Characteristics of a Couple:

1. The resultant force of a couple is zero.

2. The moment of a couple is the product of one of the forces and the perpendicular distance between their lines of action.

3. The moment of a couple is the same for all points in the plane of the couple.

A force can be resolved into another force and a couple.

10KN 10KN

10KN

10KN 2m 2m

20KN-m

Example 19: Assuming clockwise moments to be positive,

compute for the value of moment at point A.

80lb 80lb

100lb

100lb

100lb

200lb

A 1ft

1ft

Example 20: Replace the System of

forces acting on the

frame by a resultant,

R, acting at Point A

and a couple acting horizontally through B and C.

1 ft

3 ft

4 ft

2 ft

A

B

C

20 lb

30 lb 60 lb

Example 21: A couple consists of two vertical forces of 60 lb each.

One force acts up through A

and the other acts down

through D. Transform

the couple into an

equivalent couple

having horizontal

forces acting through

E and F.

A

B

C

D

E

F

G

2 in 4 in

3 in

Example 22: The three-step pulley shown in the figure is subjected

to the given couples. Compute the value

of the resultant couple.

Also determine the

forces acting at the rim

of the middle pulley

that are required to

balance the given system.

40 lb

40 lb

30 lb

30 lb

60 lb 60 lb

8’’ 12’’

16’’