Engineering Mechanics: Vector Components and Resultants, Slides of Mechanical Engineering

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BASIC SCIENCE THEORY

APPLIED SCIENCE

ENGINEERING

TECHNOLOGY

ENGINEERING MECHANICS

Course Description

• Basic Engineering Mechanics ( 4, 24 lectures )
• ( Divided in two parts, 12

lectures each )

• PART I ( S. Shabbir Ahmad ) Important vector

quantities; fundamental units; moments and couples;

resultants of forces and couples; laws of equilibrium;

free-body diagrams; tensile, compressive & shear stress

and strain; Hooke’s law; stress-strain relationship in 3D;

strain energy; plane stress and strain; principal stresses

and strains; Mohr’s circle for stress and strain; simple

introduction to strain gauges and photoelasticity;

Recommended Books

• Mechanics of Materials by F.P. Beer &

E.R. Johnston

• Mechanics of Engineering Materials by

P.P. Benham & R.J. Crawford

• Mechanics of Solids & Strength of

Materials by F.V. Warnock

• Engineering Mechanics by Timoshenko &

Young

Important Vector quantities

• Vector quantities.
• Scalar quantities.
• Distance,Mass,Speed,Charge,
• Temperature,Density,Volume,Force,
• Displacement,Velocity,Acceleration,
• Momentum,Weight,Thrust.

THE TAIL TRIP-TIP METHOD

• 1. Draw the vector to scale.
• 2. Draw the second vector to scale placing its

tail at the tip of the first vector. Make sure the

direction is correct.

• 3 Repeat (1 and 2) until all the vectors are

drawn.

• 4. An arrow drawn from the tail of the first

vector to the tip of the last to represent the

resultant of the sum.

Co-linear vectors

Perpendicular Vector

• tail to tip method.

R  a b

a

b

tan

_ 1

2 Y

2

B  BX B

VECTOR COMPONENTS

vector can be expressed as the sum of two perpendicular vectors.

vector components.

NOTE Bx and By perpendicular vector components of B. parallel to the x and y axes respectively.

B = Bx + By

Bx = B cos  and By = B sin 

Also it should be clear that

ADDITION OF VECTORS USING THE COMPONENT METHOD

• For each vector to be added resolve them into

their x & y components

• ( i.e. positive or negative x direction and

positive or negative y direction)

• Find the algebraic sum of the x components,

which represents Rx, the X component of the

resultant vector. Then find Ry by summing the y

components. This represents the Y component

of the resultant vector.

Using Pythagoras’ Theorem with Rx and Ry,determine the

magnitude of the resultant vector R.

y

x

R  R  R

Find the angle that specifies the direction of the resultant vector by using one of the trigonomic ratios.

x

_ 1 y

R

R

tan

R  R^2 x R^2 y   2. 22  2. 042  3. 00 m

Find the resultant of the three displacement vectors. The magnitudes of the

vectors are, A= 5m, B = 5m, and C = 4m.

Vector x component y component

A ax = -5 cos 20  = -4.7 m ay = 5 sin 20  = 1.71 m B bx = 5 cos 60  = 2.5 m by = 5 sin 60  = 4.33 m C cx = 4 cos 90  = 0 m cy = -4 sin 90  = -4 m

R Rx = ax + bx + cx = -2.2m Ry = ay + by + cy = 2.04m

= tan-1^ (Ry/Rx) = tan-1^ (2.04/-2.20) = 42.8 The resultant then is 3.00 m ; 43  North of West

D  D^2 x D^2 y  6. 152  3. 112  6. 89 km

Question 4 A sailboat race consists of four legs defined by the displacement vectors A, B, C & D as shown in the diagram. The magnitude of the first vectors A = 3.2 km, B = 5.1 km and C = 4.8 km. The finish line of the course coincides with the starting line. Find D and .

Vector x component y component A ax = 3.2 cos 40 = 2.45 ay = 3.2 sin 40 = 2. B bx =  5.1 cos 35 = 4.18 by = 5.1 sin 35 = 2. C cx = 4.8 cos 23 = 4.42 cy = 4.8 sin 23 = 1. dx =? dy =? R Rx = ax + bx + cx+ dx = 0 Ry = ay + by + cy + dy = 0 dx = 6.15 km dy =  3.11km

 = tan-1^ (Dy / Dx) = tan-1^ (3.11 / 6.15) = 26.8 S of E The resultant then is 6.89 km; 26.8  South of East