The branch of science which deals and records the behavior of a certain body when it is either is in motion or at rest is known as engineering mechanics. This field of science is divided into 2 parts.
In case of dynamics, it usually deals with the body and studies about it when it is in motion.
And when we are talking about statics, this branch of science deals with bodies and study about them when they are at rest.
Coming back to dynamics, it is again divided into 2 parts. They are:
It is a branch of classical mechanics which studies and is concerned with the relationship between motions of particle bodies. It also deals with the reasons which cause such motions, i.e., torque and forces.
Kinematics defines the field of science which studies a body which is in motion but does not know the reason which is the cause of this motion.
If stated in simplified words, in engineering mechanics, dynamics means body in motion and the various forces which cause it, and statics means equilibrium of a body at rest.
There are few important terms that are important in mechanics. Those are as follows:
- Scalar quantity
Any quantity whose complete specification is based on magnitude is known as magnitude. Some of the widely used scalar quantities are:
- Vector quantity
Any quantity which is specified fully by direction and magnitude and has a one dimensional measurement like weight or temperature is known as avector quantity.Depiction of vector quantity in figurative aspect is shown as a straight line from one point to another with an arrow between that line. Length of that straight line illustrates magnitude whereas the arrow in it shows direction of that vector. Some of the widely used vector quantities are:
Possessing an indefinite and extremely small volume of a body which has a negligible dimension is known as a particle. In this case, there is an assumption that mass of the body is concentrated at a certain point. Therefore, on the basis of this information, particle is believed to be that point,and the mass of it are concentrated at that point of particle.
- Law of Triangle of Forces
According to this law, when 2 forces act a point, which are direction and magnitude of a triangle (2 adjacent sides), that triangle’s closing side is taken in reverse order. That order of equilibrium depicts the resultant of both the forces representing direction and magnitude.
The figure representing the law of triangle of forces is in a slanted diamond pattern with the angle between the forces R and P as. The other angle depicts the angle between the forces Q and P.
- Law of parallelogram of forces
When 2 forces actat a certain point in a plane, to discern the result of those forces,we take the help of law of parallelogram of forces. As per the definition in engineering mechanics, “2 forces which are depicted by direction and magnitude are acting at a certain point by 2 adjacent sides of a parallelogram. Its outcome is discerned by direction and magnitude of the parallelogram’s diagonal section which passes through that point.”
Suppose we take a figure of a parallelogram which has 2 forces with names P and Q. both the forces are joined at a point which is depicted by O. the horizontal line depicting the force P represents magnitude with the direction from point O to A. similar to this, an angled vertical line which depicts the force Qrepresents magnitude and is towards the direction OQ.
Angle between the forces P and Q is represented as ‘.’When these 2 forces pass through O, which is the diagonal of the parallelogram, OP and OQ are the adjacent sides of it. The outcome of these 2 forces is acquired in magnitude.
After the creation of the parallelogram, the other point where A and B joins (opposite of O) is named C. resultant of both these forces which is illustrated in the direction from O to C is named as R.
As we have already constructed the diagram of aparallelogram and have named the forces, we will find out the magnitude of resultant.
Magnitude of Resultant (R)
Taking the help of the drawn parallelogram with different names in its various point, we will find the magnitude of the resultant which is depicted by R.
We will draw a horizontal line from A and a vertical line from G downwards so that it joins at a point of their cross section E. this will form a right angle triangle with points AD and CD.
Now, let angle between P and Q be ‘’.this angle will be some what like ∠AOB.
∠DAC = ∠AOB
This depicts corresponding angles.
In case of the portion of the parallelogram with OACB as its 4 points, you will find AC to be equal and parallel to OB. Therefore, aforce parallel to AC will be Q.
And when it is about triangle ACD,
CD =AC sin = Q sin
AD = AC cos = Q cos
In case of triangle OCD,
OC2 = OD2 + DC2
But when equating for OC,
OC = R
OD = OA +AD = P + Q cos
DC =Q sin
The next step for R will be,
R 2 = (P + Q cos )2 + (Q sin )2
= p2 + Q2 cos2 a+ 2PQ cos < X+ Q2 sin2α
= p2 + Q2 (cos2α+ sin2α)+ 2PQ cos α
= P2 + Q2 + 2PQ cos α
So finally, we will find,
R = √ p2 + Q2 + 2PQ cos α (equation 1.1)
This equation describes the magnitude of R (resultant force).
After finding the equation for the magnitude of the resultant, we will find the direction of it.
Direction of Resultant
As per the figure of aparallelogram, let us consider the angle made by OA to be θ.
After this, we will find the value of θ from OCD (triangle)
Tan θ = CD / OD = Q sin α / P + Q cos α
θ = tan -1 (Q sin α / P + Q cos α) (equation 1.2)
This value gives the direction of R.
As we have seen that the resultant is found out with the help of (tan) rule, we can also find the direction taking the help of sine rule.
According to this rule, incase of triangle OAC,
OC = R
AC = Q
OA = P
∠OACis (180 – α)
Now the value of ∠ACO is,
= 180- [θ + 180 – α]
= [(α – θ)]
On calculating with the help of sin θ,
sin θ / AC = sin (180 –α) OC = sin (α – θ) / OA
sin θ / Q = sin (180 – α) / R = sin (α – θ) / P
Now there are 2 cases that are very important.
You are already aware of the 2 forces of the parallelogram which are acting at an angle and are equal.
According to this information, direction and magnitude of a resultant are showcased in equational form. So the equation in this case is,
R = √(P2 + Q2 + 2PQ cos α)
= √(P2 +P2 + 2P x P x cos α)
= √(2P2 + 2P2 cos α)
= √(2P2 (1 + cos α))
= √(2P2 x 2 cos2 α/2)
= √(4P2 x cos2 α/2)
= 2P cos α/2
On equating further,
θ = tan -1((Q sinα)/(P+Q cosα ))
= tan -1((P sinα)/(P+P cosα ))
= tan -1(P sinα)/(P (1+cos〖 α〗))
= tan -1sinα/(1+cos〖 α〗 )
= tan -1(〖2 sin〗〖α/2〗 cos〖α/2〗)/(2 cos2〖α/2〗 )
= tan -1sin〖α/2〗/cos〖α/2〗
= tan -1(tan〖α/2〗 )
In the second case, at right angles, both forces P and Q act together. When this happens, the angle between the 2 which is depicted by ‘α’ is 90°.
With the help of previously found equation (1.1), we can easily get the resultant of the parallelogram’s magnitude. Its calculation will stand at,
R = √(P2 + Q2 + 2PQ cos α)
= √(P2 + Q2 + 2PQ cos 90°)
= √(P2 + Q2)
This is because the value of cos 90° is zero.
Now with the help of equation (1.2), we will find the calculation of resultant’s direction. In this case, it stands to be,
θ = tan -1 (Q sin α / P + Q sin α)
= tan -1 (Q sin 90° / P + Q cos 90°)
= tan -1 Q /P
It is with the help of both the stated cases above;we can come to a conclusion that it is not mandatory for any of the forces to be on the x axis.According to the equations and the figure created with the assistance of law of triangle of forces, it is clear that the forces P and Q can be in any direction.From thesecases, it is also evident that resultant of the 2 forces which are ‘α’ is equation (1.1), i.e.,
R = √ p2 + Q2 + 2PQ cos α
Now for obtaining the direction of that resultant, we need to take the help of equation (1.2), i.e.,
θ = tan -1 (Q sin α / P + Q cos α)
But the resultant angle made with the help of the direction of force P will be angle ‘e’ in this respect.
6. Lami’s theorem
Lami’s theorem is stated as an equation that interconnects 3 forces which keep a body in static equilibrium. Three forces in this respect are:
A. Concurrent forces
B. Coplanar forces
C. Non-collinear forces
As per the working of these forces, when these acts at a certain point which is in equilibrium, each of these forces become proportional to the sine of angle between remaining 2 forces.
To understand the concept clearly, let us consider 3 forces facing towards different directions. Also, let us give a figurative name of these 3 forces as P, Q and R.Their conjoining point is O. all the forces at that fixed point are in equilibrium.
After a figure is created with the help of this information, we can see the 3 differentangles formed with each other.
1. Angle between 2 forces R and P is represented byγ
2. Angle between 2 forces Q and R is represented byβ
3. Angle between 2 forces P and Q is represented byα
As it is already stated above in Lami’s theorem, P is directly proportional to the sine angle between sin β (2 forces Q and R). as per this information,
P / sin β is found constant
R / sin α is constant
Q / sin γ is also constant
P / sin β = R / sin α = Q / sin γ
This equation gave the value to Lami’s theorem. But in initial case, the proof of this equation being genuine was being questioned upon. So, an equational proof was also given of this theorem.
Proof of Lame’s Theorem
To produce Lami’s theorem, all the 3 forces were constructed in such a way that it looked like a triangle. In this case, all the forces are in equilibrium and are joined at a certain point.
P and R are joined with O, P and Q are joined with M, and R and Q are joined with N.
1. The angle at O is (180 – γ)
2. Angle at N is (180 – β)
3. Angle at M is (180 – α)
It can also be depicted as,
P / sin (180 – β) = Q / sin (180 – γ) = R / sin (180 – α)
Or its representation can be,
P / sin β = Q / sin γ = R / sin α
Hence the theorem is proved.
An important fact about this Lami’s theorem is that its forces act either towards their fixed point or away from it.
Links of Previous Main Topic:-
- Introduction to statics
- Introduction to vector algebra
- Two dimensional force systems
- Introduction concept of equilibrium of rigid body
- Friction introduction
- Introduction about distributed forces
- Area moments of inertia in rectangular and polar coordinates
- Mass moment of inertia introduction
- Work done by force
- Kinematics of particles
- Position vector velocity and acceleration
- Plane kinematics of rigid bodies introduction
- Combined motion of translation and rotation
- Rectilinear motion in kinetics of particles
- Work and energy
- Linear momentum
- Force mass acceleration
- Simple stress introduction
- Normal strain
- Statically indeterminate system
- Introduction to thermodynamics
- Statement of zeroth law of thermodynamics with explanation
- Heat and work introduction
- First law of thermodynamics for a control mass closed system undergoing a cycle
- Open system and control volume
- Conversion of work into heat
- Introduction to carnot cycle
- Clausius inequality entropy and irreversibility introduction
- Ideal gas or perfect gas
- Introduction about air standard cycles
- Properties of pure substances introduction
- Vapour compression refrigeration cycle introduction
- Basic fluid mechanics and properties of fluids introduction
- Fluid statics introduction
- Manometers measurement pressure
- Fluid kinematics
- Bernoullis equation
Links of Next Mechanical Engineering Topics:-
- Units and dimensions
- Laws of mechanics in basics and statics of particles
- Lamis theorem in basics and statics of particles
- Parallelogram and triangular law of forces
- Resolution and composition of a force
- Coplanar forces
- Resultant of coplanar forces
- Equilibrium of a particle
- Equilibrium of a rigid body
- Forces in space
- Equilibrium of a particle in space
- Equivalent system of forces
- Principle of transmissibility
- Single equivalent force
- Highlights of basics and statics of particles