CH 2: STATICS OF PARTICLES

Force component Equilibrium of a particle

Chapter Objectives: To show addition of forces (resultant force) To resolve forces into their components To express force and position in Cartesian vectors. To analyse the equilibrium of forces acted on a particle

Introduction to Force Vectors

Force » the action of one body on another. » a vector. » characterised by points of application, mag

nitude and direction. »represented by P, P or P

45º

10 N

Parallelogram Diagram P + Q = Q + P P + Q ≠ P + Q P – Q = P + (-Q)

P

QO

RR

Vector Addition

AB

C

a

c

b

Law of Sinus

a/sin A = b/sin B = c/sin C

Law of Cosines

a2 = b2 + c2 – 2bc(cosA)

b2 = a2 + c2 – 2ac(cosB)

c2 = a2 + b2 – 2ab(cosC)

VECTOR OPERATION

Vector A and its negative counterpart

Scalar Multiplication and Division

Vector Addition

Parallelogram Law

Triangle construction

Vector Subtraction

or

R’ = A – B = A + (-B)

A - B

Resolution of Vector

Extend parallel lines from the head of R to form components

Sample Problem 2.1

The two forces act on bolt at A

Determine their resultant

Q = 60N

P = 40N

Solution

Q = 60N

P = 40N

Solution

Q = 60N

P = 40N

A B

C

a

c

b

Law of Cosines

a2 = b2 + c2 – 2bc(cosA)

b2 = a2 + b2 – 2ac(cosB)

c2 = a2 + b2 – 2ab(cosC)

Solution

P = 40N

Q = 60N

P = 40N

Q = 60N

Law of Sinus

a/sin A = b/sin B = c/sin C

Solution

Q = 60N

Q = 60N

P = 40N

P = 40N

Solution

Sudut A

Alternative Trigonometric Solution

We construct the right triangle BCD and compute

CD = (60N) sin 25˚ = 25.36 N

BD = (60N) cos 25˚ = 54.38 N

tan A = 25.36 N

94.38 N

R = 25.36

sin A

A=15.04˚

R=97.73N

α = 20º + A = 35.04 R̊ = 97.7 N , 35.0˚

P = 40N

Q = 60N

Q = 60N

SP 2.2

A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is a 25

kN force directed along the axis of the barge,

determine

(a)the tension in each rope given α = 45, (b) the value of α for which the tension in rope 2 is minimum.

Sample Problem No 2.2

25kN

Solution

Tension for Graphical Solution

The parallelogram law is used; the diagonal (resultant) is known to be equal 25kNand to be directed to the right. This sides are drawn paralled to the ropes.If the drawing is done to scale, we measure

T1 = 18.5 kN T2= 13.0 kN

18.30kN

25kN

12.94kN

25kN

25kN

25kN

25kN

25kN

12.5kN

21.65kN

2.10 To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that the magnitude of P is 300 N, determine

a) The required angle if the responding R of the two forces

applied at A is to be vertical

b) The corresponding magnitude of R.

Solution (a)

Using the triangle rule and the law of sines

360 N

300 N

35˚

R

Solution (b)

β= ?We need to calculate the value of βto find the value of R

Law of Sinus a/sin A = b/sin B = c/sin C

R = 513 N

360 N

300 N

35˚

R

Rectangular Components of a Force

• x- y components• Perpendicular to each other• F = Fx i + Fy j

•

• Fx = F cos θ Fy = F sin θ

Fy = Fy j

Fx = Fx i

F

O x

y

θi

j

Example 1

A force of 800N is exerted on a bold A as shown in the diagram.Determine the horizontal and vertical componentsof the force.

Fx = -F cosα= -(800 N ) cos 35º = -655N

Fy = +F sinα= +(800 N ) sin 35º = +459N

The vector components of F are thus

Fx = -(655 N) i Fy = +(495 N) j

And we may write F in the form

F = - (655 N) i + (459 N) j

Example 　 2

A man pulls with a force of 300N on a rope attached to a building as shown in the picture.What are the horizontal and vertical components of the force exerted by the rope at point A?

Fx = + (300N) cos α Fy = - (300N) sin α

Observing that AB = 10m

cos α = 8m = 8m = 4 AB 10m 5

sin α = 6m = 6m = 3 AB 10m 5

We thus obtain

Fx = + (300N) = +240 N 45

Fy = - (300N) = -180 N 35

We write

F = + (240N) i - (180 N) j

tan θ= Fx Fy

(a)

(b) F =

(2.9)

Example 　 3

A force F = (3.1 kN)i + (6.7 kN)j is applied to bolt A.Determine the magnitude of the force and the angle θit forms with the horizontal.

6.7

Solution

6.7

First we draw the diagram showing the tworectangular components of the force and the angle θ.From Eq.(2.9), we write

Using the formula given before,

Fx = F cos θ Fy = F sin θ

θ= 65.17˚

Addition of Cartesian Vectors

R = P + Q Rxi + Ryj = Pxi + Pyj + Qxi + Qyj Rx= Px+ Qx Ry= Py + Qy Rx = Σ Fx Ry = Σ Fy

P

S

Q

O

R

O

Pyj

Syj

Sxi

Qyj

Pxi

Qxi Rxi

Ryi

2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS

How we calculate the force ?

2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS

SAMPLE PROBLEM 2.3

F2 = 80 N F1 = 150 N

F4 = 100 N

F3 = 110 N

Four forces act on A as shown.Determine the resultant of the forces on the bolt.

Solution

F2 = 80 N F1 = 150 N

F4 = 100 N

F3 = 110 N- F3 j

- ( F2 sin 20˚ ) i

( F2 cos 20˚ )

j

- ( F4 sin 15˚ ) j

( F4 cos 15˚ )

i

( F1 sin 30˚ ) j

( F1 cos 30˚ )

i

Solution

F2 = 80 N F1 = 150 N

F4 = 100 N

F3 = 110 N- F3 j

- ( F2 sin 20˚ ) i

( F2 cos 20˚ )

j

- ( F4 sin 15˚ ) j

( F4 cos 15˚ )

i

( F1 sin 30˚ ) j

( F1 cos 30˚ )

i

- F3 j

- ( F2 sin 20˚ ) i

( F2 cos 20˚ )

j

- ( F4 sin 15˚ ) j

( F4 cos 15˚ )

i

( F1 sin 30˚ ) j

( F1 cos 30˚ )

i

Ry = (14.3 N) j Rx = (199.1 N) i

PROBLEMS

2.21 Determine the x and y component of eachof the forces shown.

Solution

Solution

Problems 2.23

1.2 m 1.4 m

2.3 m

1.5 m

2.0 m

1800 N

950N900N

Determine the x and y components of each of the forces shown.

Solution

2.59m2.69m

2.5 m

Solution

2.59m2.69m

2.5 m

2.9 EQUILIBRIUM OF A PARTICLE

• The resultant force acting on a particle is zero • R = ΣF = 0• ΣFx = 0 ΣFy = 0

The forces of 20 N acting on the line but in opposite direction, passing through point A having the same magnitude. This produces a resultant of R = 0.

A

20 N

20 N

Equilibrium of Forces

F4 = 400 N

F3 = 200 N

F2 = 173.2 F1 = 300

O

OF1 = 300

F2 = 173.2 N F3 = 200

F4 = 400 N

30

30

An equilibrium system of forces produces a closed force polygon

Example

Figure 2.27 shows four forces acting on A.In figure 2.28, the resultant of the given forcesis determined by the polygon rule. Starting from point O with F1 and arranging the forces in tip-to-tail fashion,we find that the tip of F4 coincides with thestarting point O. Thus the resultant R of the given system of forces is zero, and the particle is in equilibrium.

The closed polygon drawn in Fig 2.28 provides agraphical expression of the equilibrium of A.To express algebraically the condition for theequilibrium of a particle, we write

Resolving each force F into rectangular components,we have

We conclude that the necessary and sufficientconditions for the equilibrium of a particle are

Returning to the particle shown in figure 2.27,we check that the equilibrium conditions are satisfied. We write

Newton’s First Law of Motion

If the resultant force acting on a particle is zero,the particle will remain at rest or will move with constant speed in a straight line.

Sample Problem 2.4 In a ship-unloading operation, a 15.6 kN automobile is supported by a

cable. A rope is tied to the cable at A and pulled in order to centre the automobile over its intended position. The angle between the cable and the vertical is 2, while the angle between the rope and the horizontal is 30. What is the tension in the rope?

15.6kN

15.6kN

TAB

TACTAC

TAB

Sample problem 2.6

As part of the design of a new sailboat, it is desired to determine the drag force which may be expected at a given speed. To do so, a model of the proposed hull is placed in a test channel and three cablesare used to keeps its bow on the centerline of the channel.Dynamometer reading indicate that for a given speed, the tension is 180N in cable ABand 270N in cable AE.

Determine the drag force exerted on the hull and the tension in cable AC.

2.13m 0.46m

1.22m

1.22m

180N

270N

2.13m 0.46m

1.22m

1.22m

180N

270N

Solution Determine of the Angles

First the angles α and βdefining the direction of cables AB and AC are determined.

tanα1.22m

2.13m= 1.75=

α = 60.26˚

tanβ1.22m

0.46m= 0.375=

β = 20.56˚TAC

TAE = 270 N

FD

TAB = 180 N β= 20.56˚

α= 60.26˚Free body diagram

Choosing the hull as a free body, we draw thefree-body diagram shown. It includes the forces exerted by the three cables an the hull, as well asthe drag force FD exerted by the flow.

TAC

TAE = 270 N

FD

TAB = 180 N β= 20.56˚

α= 60.26˚

2.13m 0.46m

1.22m

1.22m

180N

270N

Solution

Equilibrium Condition

R = TAB + TAC + TAE +

FD

We express that the hull is in the equilibriumby writing that the resultant of all forces is zero

Since more than three forces are involved, we resolved the forces into X and Y components

TAC

TAE = 270 N

FD

β= 20.56˚

α= 60.26˚

TAB = 180 N TAB = -(180N) sin 60.26˚ i + (180N) cos 60.26˚ j

TAC = TAC sin 20.56˚ i + TAC cos 20.56˚ j

TAE = - (270N) j

= -(156.3N) i + (89.3N) j

= TAC 0.3512 i + TAC 0.9363 j

FD = FD i

R = TAB + TAC + TAE +

FD

from,

(-156.3N + TAC 0.3512 + FD ) i + (89.3N + TAC 0.9363 – 270 N) j = 0

(180N) cos 60.26˚ j

- (180N) sin 60.26˚ i

- (270N) j

X

Y

(180N) cos 60.26˚ j

TAB = -(180N) sin 60.26˚ i + (180N) cos 60.26˚ j

TAC = TAC sin 20.56˚ i + TAC cos 20.56˚ j

TAE = - (270N) j

= -(156.3N) i + (89.3N) j

= TAC 0.3512 i + TAC 0.9363 j

FD = FD i

R = TAB + TAC + TAE +

FD

from,

(-156.3N + TAC 0.3512 + FD ) i + (89.3N + TAC 0.9363 – 270 N) j = 0

: - 156.3N + TAC 0.3512 + FD = 0

89.3N + TAC 0.9363 – 270 N = 0TAC = 193 N

FD = 88.5 N

THE END