CHAPTER 3

RESULTANTS OF COPLANAR FORCE SYSTEMS

Fy

Fx

R

Fy

Fx

Fy

Fx

R

Finding a Resultant ForceParallelogram law is carried out to find the resultant force

Resultant, R = ( P1 + P2 )

VECTOR ADDITION OF NONORTHOGONAL FORCES

The screw eye is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force.

By Parallelogram Law

select 1 force triangleto analyze

SOLUTION

Step 1 - Law of Cosines to find FR

NN

NNNNFR

2136.2124226.0300002250010000

115cos1501002150100 22

Step 2 - Law of Sines to find angle θ

8.39

9063.06.212

150sin

115sin

6.212

sin

150

N

N

NN

8.54

158.39

Angle of φ relative to X-axis

Scalar Notation x and y axes are designated positive and

negative Components of forces expressed as

algebraic scalars

sin and cos FFFF

FFF

yx

yx

RESULTANT BY COMPONENTS

F2 = F2x + F2y

F1 = F1x + F1y

FRx = F1x + F2x

FRy = F1y + F2y

FRx = ΣFx

FRy = ΣFy

–Magnitude of FR can be found by Pythagorean Theorem

Rx

RyRyRxR F

FFFF 1-22 tan and

Determine x and y components of F1 and F2 acting on the boom.

NNNF

NNNF

y

x

17317330cos200

10010030sin200

1

1

By similar triangles we have

N10013

5260

N24013

12260

2

2

y

x

F

F

NNF

NF

y

x

100100

240

2

2

12

5tan 1

FRX = Σ Fx = - 100N + 240N = 140 N

FRY = Σ Fy = 173N -100N = 73 N

NFR

15873140 22 01 5.27

14073

tan

140 N

73 NFR

The link is subjected to two forces F1 and F2. Determine the magnitude and orientation of the resultant force.

N

NNFR629

8.5828.236 22

RESULTANT FORCE

9.67

8.236

8.582tan 1

N

N ORIGINAL FORCE SYSTEM

RESULTANT OF THREE OR MORE CONCURRENT FORCES

F1

XFF

1

45

F1x

F1yF1

NNF

FX

6805

48504

51

YFF

1

35

NF

FY

510335

1

OR

01 13.5334

tan

θ=90-α=36.870

F1x=850Cos 36.87 = 680N

F1y=850Sin 36.87 = 510N

F2

F2y

F2x

30

F2x= -625sin30 = -312.5N625N

F2y= -625cos30 = -541N 750

45

F3

F3x

F3y

F3x=-750cos45 = -530N

F3y=750sin45= 530NFRX = F1x + F2x +F3x = 680N – 312.5N -530N = -162.5N

FRY = F1y + F2y +F3y = -510N - 541N + 530N = -521N

22 )521()5.162( RF

= 546N

FR

θ =01 67.72

5.162521

tan

TRANSLATION VS ROTATION

TRANSLATION

ROTATION

ROTATING TRANSLATION

M = F x d

Characteristics of MomentsCharacteristics of Moments

F

Fp

Fa

If these wheel nuts must be tightened to 85Nm, what is the force F?Angle = 35 degrees, d = 420mm.

From M = F * d, then F = M / d

Perpendicular distance = d * cos (35)

F = 85 / (0.42 * cos(35))   = 247.06 N

Example - MomentExample - Moment

A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at 0. Determine (a) the moment of 100-lb force about 0.(b) the magnitude of the horizontal force applied at A which will create the same moment about 0.(c ) the smallest force applied at A which will create same moment (d) how far from the shaft a 240 lb vertical force must act to create the same moment about 0. (e) whether any one of the forces obtain in parts (a), (b), (c ) and (d) is equivalent to the original force.

Example - MomentExample - Moment

(a) The perpendicular distance from ) to the line of action of 100-lb force is

0

24 in. cos 60 12 in.

M 100 lb 12 in. 1200 lb-in

d

Fd

The magnitude of the moment about 0

The force rotates the lever in clockwise about 0 and M0 is perpendicular to the plane.

Example - MomentExample - Moment

(b)Horizontal force

00

24 in. sin 60 20.8 in.

MM

1200 lb-in57.7 lb

20.8 in.

d

Fd Fd

F

Since the moment about 0 is 1200 lb-in the resulting F

Example - MomentExample - Moment

(c) Smallest Force, since M=Fd, the smallest value of F occurs when d is a maximum. It will be perpendicular to 0A

00

24 in.

MM

1200 lb-in50 lb

24 in.

d

Fd Fd

F

Example - MomentExample - Moment

(d)A 240-lb vertical force In this case the force is given determine the distance

00

MM

1200 lb-in5.0 in.

240 lb

0B cos 60 5.0 in. 0B 10.0 in.

Fd dF

d

Example - MomentExample - Moment

(e) None of the force in parts b, c, and d is equivalent of original 100-lb force. Although they have the same moment about 0, they have different x and y components .

ExampleExample

A vertical force P of magnitude 60 lb is applied to the crank at A. Knowing that = 75o, determine the moment P alone each of the coordinate axes.

75

60 lb

d

d = 8 cos 15 =7.727

8

T = 60lb x 7.727 in = 463.6 in-lb

Fp =60Sin75 = 57.96 lb

Fa = 60cos75 = 15.53 lb

Tx = 57.96 lb x 8 in = 463.6 in-lb

T2 = 15.53 lb x 5 in = 77.65 ft-lb

75

60 lbFp

Fa

75

The mechanism shown is used to raise a crate of supplies from a ship's hold. The crate

has total mass 56.0 . A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.310 . The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius 0.120 , the cylinder turns, and the crate is raised.

What magnitude of the force applied tangentially to the rotating crank is required to

raise the crate with an acceleration of 0.750 ? (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

Varigon’s TheoremVarigon’s Theorem

As with the summation of force combining to get resultant force

Similar resultant comes from the addition of moments

1 2 nR F F F ��������������������������������������������������������

0 R 1 1 2 2 n nM R d F d F d F d ����������������������������������������������������������������������

Moment of F1

Line of action of F1 passes through Point O.No moment is generated

Moment of F2 75 lbs

Moment of F3 60 lbs

Sum of All Moments about O

RESULTANTS OF PARALLEL FORCES

FOR MAGNITUDE

FOR LOCATION

100 lb

2

200 lb

7

A B

FR = -100lb + -200lb = -300lb

x FR

ΣMA= -100lb x 2ft + -200lb x 7 ft = -200 lb-ft - 1400 lb-ft = -1600 lb-ft

MR=ΣMA AND MR = FR x X

MA

-300lb x (X) = -1600 lb-ft

X = -1600/-300 = 5.3 ft

ΣFy = Resultant = -800lb + 600lb – 1200lb -400lb = -1800lb

Sum MA = 1800lb( x) = -600lb(3ft) + 1200lb(5ft) + 400lb(9ft) = -1800lb-ft + 6000lb-ft + 3600lb-ft MA = 7800lb-ft

x = 7800lb-ft / 1800lb = 4.33ft

DISTRIBUTED LOADS

UNIFORMLY DISTRIBUTED LINE LOADS

Brimham Rocks, North Yorkshire England

NONUNIFORMLY DISTRIBUTED LINE LOADS

b

centroid is two thirds away fromthe vertex and 1/3 away from the right angle.

(third from the right)

magnitude of distributed load comes fromarea of triangle = ½ b x h

Location of Concentrated load will be 1/3 the base length away from right angle

Value of Resultant force fromΣF

=1/3(4.5m= 1.5 m

Sum the moments about A

Solve for xbar

TRAPEZOID

Magnitude of Resultants

Lines of Action

Resultant of Dist. Loads

Resultant Point

HYDROSTATIC FORCES

calculate pressure at depth

By Pascal’s Law

P = F/A or F = PA

The distributed load looks like this

Calculated Load Intensity

Resultant

Location of Resultant

FORCE COUPLES

T=F X 2d

d d

F

F

MR

Since the 2 forces form a couple

Magnitude is the same at any point in plane

About O

About C

About B

or, I could just find theresultant of the 2 couples

Compute X resultant

Compute Y resultant

Sketch components and resultant

Compute resultant

Compute Angle