ME 101: Engineering Mechanics

A. Narayana Reddy

Department of Mechanical Engineering,Indian Institute of Technology Guwahati,

Guwahati - 781039, India.

Vectors

ScalarScalars possess magnitude but not direction. Any scalar can be represented bya single real number. Examples: mass, volume, energy, temperature.

VectorVectors possess magnitude and direction and also the addition followsparallelogram law. Examples: force, velocity, acceleration.

Classification of vectorsFixed vectors have well defined points of application that cannot bechanged without affecting an analysis.

Free vectors may be freely moved in space without changing their effecton an analysis.

Sliding vectors may be applied anywhere along their line of actionwithout affecting an analysis.

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Algebra of vectors

Equality of vectorsThe vectors are said to be equal if they have same magnitude and direction.

Negative vectorNegative vector of a given vector has the same magnitude and the oppositedirection.

Addition of vectors

A Q

RP

Parallelogram Law

Q

RP

Triangle rule for vector additionR = P + Q

Law of cosines,R2 = P2+Q22PQ cos Law of sines,sinP

=sinQ

=sin R

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Vector addition

Concurrent forces

Scalar mutiplication

Polygon rule

Repeated application of triangle rule

The triangle rule should be applied repeatedly toadd three or more vectors.

Polygon rule is applied to three are more vectors.

Repeated application of the triangle rule isequivalent to polygon rule.

Vector addition is associative,

~P+ ~Q+ ~S = (~P+ ~Q) + ~S = ~P+ (~Q+ ~S).

Mutiplication of vector with positive scalar altersthe magnitude.

Multiplication of vector with negative scalaralters the magnitude and reverse the direction.

A set of forces pass through the same point iscalled concurrent forces.

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Example 1

A barge is pulled by two tugboats. If the resultantof the forces exerted by the tugboats is 5000 Ndirected along the axis of the barge, determinea) the tension in each of the ropes for = 45,b) the value of for which the tension in rope 2 isa minimum.

Solution:

Part - a

Part - b

Part - a: Using triangle rule and law of sines, we can writeT1

sin 45=

T2sin 30

=5000 Nsin 105

T1 = 3660 N,T2 = 2590 NPart - b: The minimum tension in rope 2 occurs when T1and T2 are perpendicular.T1 = (5000 N) sin 30 = 2500 NT2 = (5000 N) cos 30 = 4330 N = 90 30 = 60.

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Rectangular components of vector

Fx = F cos x, Fy = F cos y, and Fz = F cos z.

~F = Fx~i+ Fy~j+ Fz~k

= F(cos x~i+ cos y~j+ cos z~k)

= F~

~ is a unit vector. ~ = cos x~i+ cos y~j+ cos z~k.

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Example of force components

A

B

80 m 40 m

30 m

The tension in the guy wire is 2500 N.Determine:a) components Fx, Fy, Fz of the forceacting on the bolt A,b) the angles x, y, z defining thedirection of the force.

A

B Solution: Let ~ be the unit vector pointing from A to

B. Then vector ~ =~ABAB

= 0.424~i+ 0.848~j+ 0.31~kas the vector ~AB = 40~i+ 80~j+ 30~k.Force, ~F = F~ = 2500~ = 1060~i+ 2120~j+ 795~kcos x = 0.424, cos y = 0.848, cos z = 0.31.x = 115.1, y = 32.0 and z = 71.5

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Example 3

Example - 3:The two forces act on a bolt at A. Determinetheir resultant.

Solution approaches:1 Graphical solution by constructing parallelogram. The direction of

forces should be taken from the given data whereas length of arrowshould be proportional to magnitude.

2 Trigonometric solution can be obtained by triangle rule. The law ofcosines and law of sines should be applied to get magnitude anddirection, respectively.

3 First the given forces should be resolved into components alongcoordinate axes. Later the corresponding components should be added toget resultant force.

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Solution

i) Graphical slution:R = 98 N = 35.

ii) Trigonometric solution:Law of cosines: R2 = P2 + Q2 2PQ cosB = R = 97.73 N.Law of sines:

sinAQ

=sinBR

= sinA = (sinB)QR

.

A = 15.04 and = 20 + A = 35.04.iii) Solution using components:~P = 40[cos(20)~i+ sin(20)~j] = 37.58~i+ 13.68~j.~Q = 60[cos(45)~i+ sin(45)~j] = 42.43~i+ 42.43~j.~R = ~P+ ~Q = ~R = 80.01~i+ 56.10~j.R = 97.72 and = 35.03.

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