Transcript STATICS LECTURE

STATICS
TOPIC
–
FORCES ACTING AT A
POINT
RESULTANT AND
COMPONENTS
RESULTANT AND COMPONENTS
If two or more forces act upon a rigid body and if a
single force can be found such that the effect of it
upon the body is same as that of all the forces
taken together, then the single force is called the
RESULTANT of the forces and the given forces
themselves are called COMPONENT OF FORCES.
NOTE:
Forces are said to be in
EQUILLIBRIUM if there is no
resultant force.
PARALLELOGRAM
LAW OF FORCES
PARALLELOGRAM LAW OF FORCES
If two forces acting at a point be represented in
magnitude and direction by the two sides of a
parallelogram through the point of application ,
their resultant will be completely represented by
the diagonal of the parallelogram through that
point.
B
Q
O
C
R
P
A
OA + OB = OC
MAGNITUDE AND DIRECTION OF
RESULTANT
Let two given forces P and Q acting at at an angle α be
represented in magnitude and direction by OA and OB resp..
Complete the parallelogram OABC then the resultant R is
represented in magnitude and direction by the diagonal
OC.Draw CD perpendicular to OA as in fig.
B
C
R
Q
α
180-α
θ
O
α
P
A
OA + OB = OC
Since AC and OB are the parallel and equal
,therefore AC will also represent the force Q.
Also since angle AOB=α, thus angle CAD=α and
180- α in fig.
OD=OA+AD
= OA+AC+AC.AD/AC=P+Q cosα
And CD=AC sinα=Q sinα
OD=OA-DA
= P-Q cos (180-α )
= P+Q cos α
CD=Q sinα
OD=P+Q cosα andCD=Qsinα
Now in right angle OCD
OC²=OD²+CD²
R=(P+QCOSα)+(Qsinα)
R=sqrt(P²+Q²+2Pqcosα)
and
tanθ=CD/OD
=Qsinα/(P+Qcosα)
DEDUCTIONS
COR.1
if ø be the angle which the
resultant makes with OB
,then
tan ø = P sin α/(Q+P cos α)
COR.2
 If the two forces P and Q are perpendicular to
one another i.e. if α=π/2
 Then
R=√(P²+Q²+2PQcosπ/2)
R=√(P²+Q²)
And
Tanθ=Qsin π /2/(P+Qcos π /2)=Q/P
COR.3
 If P=Q i.e. if two forces are equal then
R=√(P²+P²+2P²cosα)
=√(2P²(1+cosα))
=√(2P²(2cos²α/2))
therefore R=2Pcosα/2
And tanθ=Psinα/(P+Pcosα)=sinα/(1+cosα)
=tanα/2
θ =α/2
COR 4.
 If P>Q, then P+Qcosα>Q+Qcosα
i.e (P+Qcosα)/(Qsinα)>(Q+Qcosα)/(Qsinα)
Dividing throughout by Qsinα,we get
Qsinα/(P+Qcosα)<sinα/(1+cosα)
Tanθ<(2sinα/2.cosα/2)/2cosα/2
Tanθ<tanα/2
θ<α/2
COR. 5
 Maximum value of resultant:
We have
R²=P²+Q²+2PQcosα……………………………(1)
From (1)
R is maximum when cosα is maximum .But
maximum value of cosα =1
i.e. Whenα =0˚
Therefore R²=P²+Q²+2PQ=(P+Q)²
R=P+Q
COR.6
 Minimum value of resultant:
We have R²=P²+Q²+2PQcosα…………..(1)
From (1), R is minimum when cosα is minimum.
But minimum value of cosα =-1
i.e. α =180˚
Therefore ,R²=P²+Q²+2PQ(-1)
R²=(P-Q)²
Therefore ,R=P-Q
EXAMPLE :1
 Find the magnitude and direction of the
resultant of two forces of magnitudes 12N
and 14N ,acting at a point and inclined to
each other at an angle of 45 ˚.
B
C
14
R
45˚
θ
O
12
A
SOLUTION
 Let P=12N , Q=14N and let R be the resultant acting at O
,making an angle θ with P (=12N) .also α = 45˚
 Therefore R=√(P²+Q²+2PQcosα)
=√((12)²+(14)²+2.12.14.cos45˚)
=√(144+196+168.√2)
=√(144+196+168.(1.414))
=√577.552
=24.03N
Also
Tanθ=Qsinα/(P+Qsinα)
=14.sin45˚/(12+14.sin45˚)
=14.(1/√2)/(12+14.(1/√2))
=14/(12.√2+14)
=0.45
Therefore
θ=tan ‫־‬¹(0.45)
Thus the resuLtant is of magnitude 24.03N and
makes an angle of tan‫־‬¹ (0.45)with the direction
Of force 12N.
ASSIGNMENT
1.Two forces of magnitudes 8N and
6N acts at a point and the angle
between them is 60 ˚.find the
magnitude and direction of their
resultant ?
2. Find the angle between two
forces P,P when the square of their
resultant is equal to (2-√3) times
their product?
3.Two forces acting at a point are
such that if the direction of one
is reversed the direction of
resultant is turned through a right
angle prove that the forces must be
equal in magnitude?
4.Two forces P and Q have a
resultant R .if the force P be
increased then the new resultant
bisect the angle between R and P
.find the increase in P?
5.When two equal forces are
inclined at an angle 2α their
resultant is twice as great as when
they are inclined at an angle 2β .
Prove that cosα=2cosβ ?
6.The resultant R of forces P and Q
makes an angle 2θ with the line of
action P .P is now replaced by P+R
, Q remaining unchanged .show that
the resultant makes an angle θ with
P?
TEST
NOTE: do any
two ?
1.To find the magnitude and
direction of the resultant of two
forces acting at a point ?
2.Two forces P and 2P acts on a
particle if the first be doubled
and the second be increased by 10
kg weight . The direction of the
resultant is unaltered . Find the
value of P?
3. The resultant of forces P and Q is R
.if Q be doubled ,R is doubled, if Q be
reversed
, R is again doubled . Show
that
P²:Q²:R²=2:3:2
or
P:Q:R=√2:√3:√2
4. two forces P+Q and P-Q makes an angle
2α with one another and their resultant
makes an angle ‘θ’with the bisector of
angle between them .show that
Ptanθ=Qtanα?