engr221 lecture 3

Concurrent Force Systems(II)Concurrent Force Systems(II)

ENGR 221

January 22, 2003

Lecture GoalsLecture Goals

• 2.6 Rectangular Components of a Force

• 2.7 Resultants by Rectangular Components

• 3.2 Free-Body Diagrams

• 3.3 Equilibrium of a particle

Two Dimensional ResultantsTwo Dimensional Resultants The vector components are written as

x yF F i F j

Fx

Fy

F

where,

i - the unit vector in the x direction

j - the unit vector in the y direction

Two Dimensional VectorsTwo Dimensional Vectors Vector Components:In two dimensions, a force can be described using a magnitude |F| and single angle, The components of the vector are Fx and Fy

2 2x y

xy1

yx

F

Fcostan

Fsin

F F

FF

FF

Fx

Fy

F

Two Dimensional Vector - ExampleTwo Dimensional Vector - Example

Given two points(x1,y1) = (4,5) and (x2,y2) = (16, 10), determine the vector from 1 to 2. Determine the components of the vector.

x 2 1

y 2 1

16 4 12

10 5 5

F x x

F y y

Vector can be written as:

x yF

12 5

F i F j

i j

Two Dimensional Vector - ExampleTwo Dimensional Vector - Example

y1

x

1

tan

5tan 0.3948 rad

12

22.62

F

F

2 2x y

2 2

F

12 5 13

F F

The magnitude and angle are

Three Dimensional VectorsThree Dimensional Vectors The vector components are written as

x y zF F i F j F k

where, i - the unit vector in the x direction j - the unit vector in the y direction k - the unit vector in the z direction

Three Dimensional VectorsThree Dimensional Vectors Vector Components:In two dimensions, a force can be described using a magnitude |F| and three angles,x, y, and z The components of the vector are Fx, Fy, and Fz.

2 2 2x y zF F F F

Three Dimensional Three Dimensional VectorsVectors

Vector Components:The three angles,x, y, and z are defined as:

xx

yy

zz

cosF

cosF

cosF

F

F

F


Vector Components:These vector cosines are

x x

y y

z z

F cos

F cos

F cos

F

F

F

x y zF F i F j F k


Vector Components:Substitute into the vector

x y z

x y z

F F cos F cos F cos

F cos cos cos

i j k

i j k

The magnitude is |F| and unit vector is

x y zcos cos cosi j k


Vector Components:The unit vector can be written as:

The magnitude is |F| and unit vector is 2 2 2x y z

2 2 2x y z

1

cos cos cos 1

x y z

x y z

cos cos cosi j k

i j k

Three Dimensional Vector - ExampleThree Dimensional Vector - Example

Given two points(x1,y1,z1) = (4,5,6) and (x2,y2 ,z2) = (16, 10,12), determine the vector from 1 to 2. Determine the components of the vector.

x 2 1

y 2 1

z 2 1

16 4 12

10 5 5

12 6 6

F x x

F y y

F z z

Vector can be written as:

x y zF

12 5 6

F i F j F k

i j k


2 2 2x y z

2 2 2

F

12 5 6 14.3178

F F F

The magnitude is


The directional cosines are

xx x

yy y

zz z

12cos 0.8381 0.577 rad or 33.06

F 14.3178

5cos 0.3492 1.214 rad or 69.56

F 14.3178

6cos 0.4191 1.138 rad or 65.22

F 14.3178

F

F

F


The vector can be written as

x y zF F cos cos cos

14.3178 0.8381 0.3492 0.4191

i j k

i j k

Class - ProblemClass - Problem

Determine the magnitude and directional cosines of the vector.

A 700 820 900i j k

Why?Why?

Why do we need to a procedure to find the directional cosine or unit vector of the vector?

Force - ExampleForce - Example

A tower guy wire is anchored by means of a bolt at A. The tension in the wire is 2500 N. Determine (a) the components Fx, Fy, and Fz of the force acting on the bolt,(b) the angles x, y, and z defining the direction of the force.


The first step is to determine the coordinate system for the vector AB. If we place the forces acting on bolt A (tension). The force acts along the direction of the wire, so we length of the vector to fined the unit vector

x

y

z

40 m

80 m

30 m

d

d

d

Force - ExampleForce - ExampleThe magnitude of d is

The unit vector is

2 2 2d 40 m 80 m 30 m

94.34 m

40 m 80 m 30 md 94.34 m

94.34 m 94.34 m 94.34 m

94.34 m 0.424 0.848 0.318

0.424 0.848 0.318

i j k

i j k

i j k

Force - ExampleForce - ExampleThe magnitude of the force is 2500 N so that the force vector is

So that Fx = -1060 N, Fy = 2120 N and Fz = 795 N.

F

2500 N 0.424 0.848 0.318

1060 N 2120 N 795 N

F

i j k

i j k


The angles are

x x

y y

z z

cos 0.424 2.009 rad or 115.1

cos 0.848 0.559 rad or 32.0

cos 0.318 1.247 rad or 71.5

Resultant ForcesResultant Forces

The components of vectors are used to find the resultants acting on object. Using the unit vectors, the components of forces are

x x y y z z R F R F R F 2 2 2x y z

yx zx y z

R

cos cos cosR R R

R R R

RR R

Class - ProblemClass - Problem

Cable AB is in tension, 450 lbs, determine the components of the force exerted on the plate at A.

Equilibrium of a Particle in SpaceEquilibrium of a Particle in Space

The components of the forces in equilibrium

x y z0 =0 0F F F

Example-EquilibriumExample-Equilibrium

A 200 kg cylinder is hung by means of two cable AB and AC, which are attached at the top of a vertical wall holds the cylinder in the position shown. Determine the magnitude of P and the tension in each of the cables.


Define the coordinate system of the points and find the force vectors.

2 200 kg 9.81 m/s

1962 N

W mg j j

j

P P i


The vector AB is

x

y

z

1.2 m

12 m 2 m 10 m

8 m

d

d

d

2 2 2

1.2 m 10 m 8 m

1.2 m 10 m 8 m 12.86 m

AB i j k

AB

��


The unit vector in the AB direction is

The tension vector TAB is

AB

1.2 m 10 m 8 m

12.86 m 12.86 m 12.86 m

0.0933 0.778 0.622

ABi j k

AB

i j k

��

AB AB AB

AB AB AB0.0933 0.778 0.622

T T

T i T j T k

��


The vector AC is

x

y

z

1.2 m

12 m 2 m 10 m

10 m

d

d

d

2 2 2

1.2 m 10 m 10 m

1.2 m 10 m 10 m 14.19 m

AC i j k

AC

��


The unit vector in the AB direction is

The tension vector TAB is

AC

1.2 m 10 m 10 m

14.19 m 14.19 m 14.19 m

0.0846 0.705 0.705

ACi j k

AC

i j k

��

AC AC AC

AC AC AC0.0846 0.705 0.705

T T

T i T j T k

��


Apply the equilibrium condition.

Combine the vectors AB AC0 0F T T W P ��

AB AC

AB AC

AB AC

0

0.0933 0.0846

0.778 0.705 1962 N

0.622 0.705

F

T T P i

T T j

T T k


Break the components of the forces:

Use (3) to get a relationship for TAB and TAC

AB ACx

AB ACy

AB ACz

0 0.0933 0.0846 1

0 0.778 0.705 1962 N 2

0 0.622 0.705 3

F T T P

F T T

F T T

AB AC AC

0.7051.133

0.622T T T


Plug into (2) solve for magnitude of TAC and TAB

Use the values to solve for P

AC AC

AC

AB

0.778 1.133 0.705 1962 N

1236 N

1401 N

T T

T

T

AB AC0.0933 0.0846

0.0933 1401 N 0.0846 1236 N

235 N

P T T

Class -ProblemClass -ProblemA weight W is supported by three cables. Determine the value of W, knowing that the tension in the cable DC is 975 lb.

Class-ProblemClass-Problem

The tension force in AC is 28 kN, determine the the required values of tension in AB and AD so that the resultant force of the three forces applied at A is vertical. Determine the resultant force.

Free Body DiagramsFree Body Diagrams

The first step in solving a problem is drawing a free-body diagram (FBD). This step is the most crucial and important solving any problem. It defines weight of the body, the known external forces, and unknown external forces. It defines the constraints and the directions of the forces. If the FBD is drawn correctly the solving of the problem is trivial.


Construction of a free body diagram.

Step 1: D

Step 2:

Step 3:

Decide which body or combination of bodies are to be shown on the free-body diagram.

Prepare drawing or sketch of the outline of the isolated or free body.

Carefully trace around the boundary of the free-body and identify all the forces exerted by contacting or attracting bodies that were removed during isolation process.


Construction of a free body diagram(cont.)

Step 4: CChoose the set of coordinate axes to be used in solving the problem and indicate their directions on the free-body diagram. Place any dimensions required for solution of the problem on the diagram.

FBD - ExamplesFBD - ExamplesWhat is the free-body diagram of the weight?

FBD - ExamplesFBD - ExamplesThe diagram has the given angles and the know magnitudes.

TDA

TDB

TDC

W

FBD - ExamplesFBD - ExamplesWhat is the free-body diagram of the tower for the resultant force acting vertically?

FBD - ExamplesFBD - ExamplesThe resulting free-body diagram can solve the problem by attaching points.

TAB

TAD

TAC

R

Homework (Due 1/29/03)Homework (Due 1/29/03)

Problems:

2-55, 2-61, 2-63, 2-83, 3-3, 3-4

engr221 lecture 3

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