mechanics unit 1

8/20/2019 mechanics unit 1

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Composition and Resolution of

Forces

Unit -1


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Force

• Force is an agent which tend to cause eitherchange in motion or deformation.

• It is denoted by N (Newton) = kg.m/s2

• Effect of Force

– Change the motion of body

– Retard the motion of body

– Deform the body – Bring the body to rest/equilibrium

– Change the stress in the body


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Characteristics of Force

1. Magnitude of force (N)

2. Line of action of force (direction)3. Nature of force (pull/push)

4. Point of action of force

System of Forces

1. Coplanar forces

2. Collinear forces

3. Concurrent forces

4. Coplanar forces

5. Coplanar concurrent forces

6. Non-coplanar concurrent forces

7. Non coplanar non concurrent forces


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• Principle of independence of forces

If a number of forces are simultaneously acting on a

body, then the resultant of these forces will have the

same effect as produced by all the individual forces.

• Principle of Transmissibility of forces

The effect of an external force on a rigid body

remains unchanged even if the force is moved, along its lineof action.

= + +

ine of actionP = =

=P P P


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Resultant force

• If a body is subjected to n number of forces, acting

simultaneously, the representive single force whichproduces the same effect is called the resultant force.

• The process of determining the resultant force of a givensystem of forces on a rigid body is composition of forcesor compounding of forces.

• Methods for determining the resultant force

• Analytical Method

–Parallelogram law of forces –Resolution of force

• Method of resolution


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Analytical Method

• Parallelogram law of forces

–If two forces acting simultaneously on a body, they canbe represented by adjacent sides of parallelogram in

which the resultant may be represented in magnitude

and direction by the diagonal of the parallelogram

which passes through their point of intersection


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= 2 cos

tan = sin

cos

Cases1. If the forces are in straight line( = 0), R = P + Q

2. If the forces are perpendicular( = 0), =

3. If the forces are acting opposite in straight line ( = 180), R=P - Q

4. If the forces are equal (P=Q), R = 2P cos


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Numerical

• Two forces acting at an angle of 120o of which the higher

magnitude of force was 40N and the resultant is

perpendicular to the other force, find the other force.

F2

120oR

90o


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Graphical method

• Draw a line OA and OB to scale for forces P and Q

respectively, with angle with each other.

• Construct the parallelogram OABC by erecting parallel to

OA and OB to intersect at C.

• Draw a diagonal OC, which represent the reaction force,the measure of OC, multiplied with scale gives the

magnitude of R.

O A

CB

D

P

Q

R

• The angle AOC is the direction of

reaction R.• Extent line OA and mark a point D

such that OD CD

• OD represents the projection of the

resultant R.


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Analytical Method

• Resolution of force

– The most common method adopted for resolution of forces is

into rectangular components.

– So, as per the parallelogram law of forces, Force F can be

written as

F = Fx + Fy

– In vector form, the same can be represented as

F = Fx i + Fy j

– Considering the angle between horizontal component Fx and

F is ,

Fx = F cos =

Fy = F sin = tan−

F

Fx

Fy

i

j

x

y


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Hence, in the given system of forces acting on the body in different

directions, the resultant of the forces can be determined by

resolving the forces in two directions and obtain the vector sum ofthese components.

Mathematically, it can be determined as follows:

R = F1 + F2 + F3 +. . .= (F1x i + F1y j) + (F2x i + F2y j) + (F3x i + F3y j) + . . .

Rx i+Ry j = (F1x + F2x + F3x+ . . .)i + (F1y + F2y + F3y + . . .)j

R =

and = tan−

R is the reaction force magnitude

is the direction, represented by angle of reaction w.r.t. Horizontal


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Various situations and solutions

Fx = F sin

Fy = F cos

x

y

F

x

y

F

Fx = - F cos

Fy = - F sin

x

yF

Fx = F sin (-)

Fy = - F cos (-)

x

yF

x

yF

Fx = F cos (-)

Fy = F sin (-)


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The foces F1, F2, and F3, all of which act

on point A of the bracket, are specified

in three different ways. Determine the xand y scalar components of each of the

three forces.

F 1x = 600 cos 35° = 491 N

F1y = 600 sin 35°

= 344 N

F 2x = -500 (4/5) = - 400 N

F2y = 500 (3/5) = 300 N

= tan- 1 [0.2/0.4] = 26.6 °

F 3x = 800 sin 26.6 = 358 N

F 3y = - 800 cos 26.6 = - 716 N


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R x = 100 cos 30o + 80 cos 20o = 161.8 N

Ry = 100 sin 30o

- 80 sin 20o

= 22.6 NR = 161.8 22.6 = 163.4 N

= tan-1 (22.6/161.8) = 8o

R2 = (80)2 + (100)2 + 2(80)(100) cos 50o

tan =00 50

0+00 50

R = 163.4 N = 28o wrt F2,

i.e. 8o wrt X axis


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Moment• In addition to the tendency to move a body in the direction of its

application, a force can also tend to rotate a body about an axis.• This rotational tendency is known as the moment M of the force,

also referred to as torque.

• The moment is a vector M perpendicular to the plane of the

body.• Therefore, the magnitude of the

moment is defined as

M = F d

where d is also termed as lever armdistance, perpendicular to the line offorce.

• The concept is highly used in hinges(for doors, windows etc.)


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Varignon's Theorem states that the moment of a force about

any point is equal to the sum of the moment of the

components of the force about the same point.

Mo = Q × r + r × P = R × r Mo = Q.q + (- P.p) = R.d

M = F d = r × F =F r sin


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Additional information

Hinge support: The structural component whichsupports vertical force, horizontal force, but momentis zero.

Roller support: The structural component whichsupports vertical force, but both horizontal force andmoment is zero.

Fixed support: The structural component whichsupports all vertical force, horizontal force andmoment.


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Couple• The moment produced by two equal, opposite, and non-collinear

forces is called a couple.

• These two forces cannot be combined into a single force as their

resultant in every direction is zero, their only effect is to produce a

tendency of rotation, same as moment.

• Hence, for the above situation,

Couple M = F.(a+d) – F.a = F.d (anticlock)

where d is the distance between the forces

• This indicates that the couple is independent on the point of moment.

• In vector form, M = ra ×F + rb ×(-F) = (ra-rb )×F = r × F


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Clockwise

couple

Anti-Clockwise

couple

A couple is not affected if the forces act in a different but. parallel plane


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Translation of forces

A

B

.

.

F A

B

.

.

F

FF

d= AB

.

F

M= F d

.

=


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ProblemIn the given figure, if the force of 100 N is

replaced with force P = 400 N, causing the

same couple. determine the angle,

The magnitude of the moment due to force of

100 N is

M = 100 (100) = 104 N.mm = 10 N.m (anti-clock)

Similarly, the moment due to forces P,produce a moment of

M = 400 (40 cos ) = 1.6 cos N.m (anti-clock)

Equating both, = 51.3o


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Replace the given force by equivalent

force-couple at O80 N

Solution:

Moment at O is

M = 80 (9 sin 60o)

= 624 N.m

Force = 80 N


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Resultants

• The resultant of a system of forces is the simplest force

combination which can replace the original forces without altering

the external effect on the rigid body to which the forces are applied

R = F1+F2+F3+…. = F

Rx= Fx , Ry= Fy , =

= tan−

= tan−


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Problem• Determine the resultant {If the four forces and one

couple which act on the plate shown

Rx= 66.9 N

Ry= 132.4 N

R = 148.3 N

= 63.2o

Mo

= 140 – 50 (5) + 60 cos 45o (4) – 60 sin 45o (7)

= -237 N.m = 237 N.m (anticlock)

.OMo= 237 N.m

R = 148.3 N

= 63.2o

.OMo

= 63.2o

Final Line of resultant of R is at distance d from O,

i.e. d = 237/148.3 = 1.6m

So the line of action forms tangent making an angle 63.2o with horizontal


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End of Unit 1

mechanics unit 1

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