Laws of Mechanics
1 Newton’s law
Law I
Each body remains in its state of rest or motion uniform in direction until it is made to change this state by imposed forces.
Law II
The change of motion is proportional to the imposed driving force and occurs along a straight line in which the force acts.
Law III
To every action there is always an equal reaction: or the mutual interactions of two bodies are always equal but directed contrary.
2 Lami’s theorem
It states that, “If there forces acting at a point are in equilibrium, each force will be proportional to the sine of the angle between the other two forces.”
Suppose the three forces P, Q and Rare acting at a point 0 and they are in equilibrium as shown in Fig.
Let a = Angle between force P and Q.
β =Angle between force Q and R.
y = Angle between force R and P.
Then according to Lame’s theorem, P is proportional sine of angle between Q and R a sin β. P / sin β = constant.
Similarly Q / sin γ = constant,
R / sin a = constant P / sin β = Q / sin γ = R / sin a
Example-1: In a jib crane, the jib and the tie rod are 5 m and 4 m long resp ectively. the height of crane post in 3 m and the ti es red remains horizontal. Determine the for ces produced in jib and tie rod when a load of 2 kn in suspended at the crane head.
Solution: From figure
sin q = 3/5 = 0.6
q = 36.87o
Let P1 and P2 be the forces developed in jib and tie rod respectively. the three forces P1, P2 and W are shown in figure with the angle between the forces calculated from the given directions. The line of action of forces P1, P2 and weigh W meet at the point C, and therefore Lami’s theorem is
applicable. That gives:
∴P1/sin 270o = P2/sin 53.13o = 2/si n36.87o
P1 = 2 × sin 270 o / sin 36.87o = 2 × 1/0. = –3.33 kN
P2 = 2 × sin 53.13 o / sin 36.87o = 2 × 0.8 / 0.6 = 2.667 kN
The –ve sign indicates that the direction of force P1 is opposite to that shown in figure obviously the tie rod will be under tension and jib will in compression.
Examples-2: A string ABCD w hose extremity A is fixed has weights W1 and W2 attached to it at B and C, and passes round a smooth peg at D carrying a weight of 800 N at the free end E shown in Figure. If in a state of equilibrium, BC is horizontal and AB and CD make angles of 150o and 120o respectively wit h BC, make calculation for (a) The tension i n portion AB, BC, CD and DE of the string. (b) t he value of weights W1 and W2 (c) The load o n the peg at D
Solution: Let T1, T2, T3, T4 be the tension in segments AB, BC, CD and DE of the string.
Under equilibrium condition, T5 = T4 = 800 N
Applying Lami’s theorem at point B,
T1/sin90o = T2/sin120o = W1/sin150o
T1 = T2 sin 90o / sin 120o = 400 × 1/0.866 = 461.89 N
W1 = T2 sin 150o / sin 120o = 400 × 0.5 / 0.866 = 230.95 N
(c) Load on peg at D = T3 sin 60o + W
= 800 sin 60o + 800 = 692.82 + 800
= 1492.82 N
3 Parallelogram Law of forces
The law of parallelogram of forces is used to determine the resultant* of two forces acting at a point in a plane. It states, “If two forces, acting at a point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.”
Let two forces P and Q act at a point 0 as shown in Fig. 1.3. The force P is represented in magnitude
and direction by OA whereas the force Q is presented in magnitude and direction by OB. Let the angle between the two forces be ‘a’. The resultant of these two forces will be obtained in magnitude
and direction by the diagonal (passing through O) of the parallelogram of which OA and OB are two adjacent sides. Hence draw the parallelogram with OA and OB as adjacent sides as shown in Fig. The resultant R is represented by OC in magnitude and direction.
Magnitude of Resultant (R)
From C draw CD perpendicular to OA produced.
Let a = Angle between two forces P and Q = LAOB
Now < DAC = < LAOB (Corresponding angles)
In parallelogram OACB, AC is parallel and equal to OB .
AC=Q.
In triangle ACD,
AD = AC cos a = Q cos a and CD =AC sin a= Q sin a.
In triangle OCD,
OC2 = OD2 + DC2.
But OC = R, OD = OA +AD = P + Q cos a and DC =Q sin a
R 2 = (P + Q cos a)2 + (Q sin a)2 = p2 + Q2 cos2 a+ 2PQ cos < X+ Q2 sin2 a
= p2 + Q2 (cos2 a+ sin2 a)+ 2PQ cos a
= P2 + Q2 + 2PQ cos a
R = √ p2 + Q2 + 2PQ cos a
Direction of Resultant
Let θ = Angle made by resultant with OA.
Then from triangle OCD,
tan θ = CD / OD = Q sin a / P + Q cos a θ = tan -1 ( Q sin a / P + Q cos a)
The direction of resultant can a lso be obtained by using sine rule [In triangle O AC, OA = P, AC = Q, OC = R, angle O AC = (180 – a), angle ACO = 180- [θ + 180 – a] = (a- θ)]
sin θ / AC = sin (180 –a) OC = sin (a – θ) / OA sin θ / Q = sin (180 – a) / R = sin (a – θ) / P
Two cases are important.
1st Case. If the two forces P and Q act at right angles, then a = 90° we get the m agnitude of resultant as
the direction of resultant is obtai ned as θ = tan -1 ( Q sin a / P + Q sin a)
= tan -1 ( Q sin 900 / P + Q cos 90 0 ) = tan -1 Q /P
2nd Case. The two forces P and Q are equal and are acting at an angle a betwee n them. Then the magnitude and direction of resul tant is given as
4 Law of Triangle of Forces
It states that, “if three fo rces acting at a point be represented in magnitude and direction by the three sides of a triangle, take n in order, they will be in equilibrium.”
5 Vectorial representation of forces