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Monday, 30 March 2020

March 30, 2020

LEVER AND ITS TYPES WITH EXAMPLES

We were discussing the projectile motion - trajectory equation, definition and formulas and we have also seen the concept of Terminal velocityapparent weight of a man in a lift and Lami's theorem in engineering mechanics with the help of our previous posts.  

Today we will be interested here to start another important topic in engineering mechanics i.e. Definition of levers and it’s types or classifications with the help of this post. 

Levers 

A lever is basically defined as machine member or a rigid bar which is hinged at one end or point. This hinge end or point will be termed as fulcrum. Lever will be free to rotate about this fulcrum. 

A lever will have a point to apply the effort and it is termed as effort arm and similarly lever will also have a point for lifting the load which is termed as load arm. 

Following figure, displayed here, indicates the lever. We can see here the fulcrum, load arm and effort arm. Load arm of a lever is also called as resistance arm. 

Therefore, there will be three main components of a lever i.e. fulcrum, effort arm and load arm. 

Types of levers 

There are basically two types of levers i.e. simple lever and compound lever as mentioned here. 

Simple lever 

The lever which consists of one bar having one fulcrum will be termed as simple lever and it is displayed here in following figure. 

We must note it here that this lever might be straight, curved or bent. Let us do some calculation for simple lever as mentioned below.
Where,
P is the effort applied on effort arm of length a
W is the load applied over the load arm of length b 

Let us take moment about the fulcrum and we will have following equation as mentioned below 

P x a = W x b

W/P = a/b 

W/P will be termed as mechanical advantage and a/b will be termed as leverage of the lever. 

After analyzing the equation of mechanical advantage, we can easily say that in order to increase the mechanical advantage, either we need to increase the length of a or we need to decrease the length of b. 

Types of simple levers 

Further, simple levers are classified in following three types as mentioned below.
  • First order simple lever or first class lever 
  • Second order simple lever or second class lever 
  • Third order simple lever or third class lever 

First order simple lever 

If the fulcrum of lever is located in between the effort and load, then such lever will be termed as first order lever.

Example of first order simple lever

Hand pump
Punching press hand wheel
Seesaw
Scissor
Rocker arm
Beam Balance 

Second order simple lever 

If the length of effort arm is more than the length of load arm, then such lever will be termed as second order simple lever. 

In second order simple lever or class 2 lever, 

Length of effort arm > Length of load arm, hence applied effort will be less in case of second order simple lever. 

Example of second order simple lever 

Dead weight safety valves 

Third order simple lever

If the length of effort arm is less than the length of load arm, then such lever will be termed as third order simple lever. We can also say that in case of third order simple lever, length of load arm will be greater than the length of effort arm. 

In third order simple lever or class 3 lever, 

Length of load arm > Length of effort arm 


Fig: First class, Second Class and Third class lever 

Example of third order simple lever 

Tongs and treadle of sewing machine 


Compound lever 

Compound lever is basically a combination of a number of simple levers. Following figure indicates the compound lever. 

Mechanical advantage or leverage will be greater in case of compound lever as compared with simple lever. 

Leverage of compound lever = Leverage of first lever + Leverage of the second lever + ……  

Fig: Compound lever 

Example of compound lever 

Platform weighing machine is the best example of a compound lever 

Therefore, we have seen here the basics of levers and its types with the help of this post. We have also secured here the information of first order, second order and third order levers. 

We have also seen here the various examples of simple lever, compound lever,  first order simple lever, second order simple lever and third order simple lever with the help of this post. 

Further we will find out another concept in engineering mechanics with the help of our next post.   

Do you have any suggestions? Please write in comment box and also drop your email id in the given mail box which is given at right hand side of page for further and continuous update from www.hkdivedi.com.  

We will find out now the in our next post.   

Reference:  

Basic engineering mechanics and strength of materials, By M M Das , M D Saikia and B M Das
Image courtesy: Google     

Also read  


Sunday, 1 March 2020

March 01, 2020

LAMI'S THEOREM IN ENGINEERING MECHANICS

We were discussing the projectile motion - trajectory equation, definition and formulas and we have also seen the concept of Terminal velocity and apparent weight of a man in a lift with the help of our previous posts.  

Now, we will be interested further to understand a very important topic i.e. Lami's theorem in engineering mechanics. 

We will find out here the statement, explanations, importance, applications and limitations of lami's theorem with the help of this post. We will also see here the Conditions to apply the Lami’s theorem. 


Lami's theorem in engineering mechanics 

Let us first see here what is the statement of Lami's theorem in engineering mechanics.
Lami’s theorem states that 

If three coplanar, concurrent and non-collinear forces are in equilibrium, the magnitude of each force will be proportional to the sine of the angle between the other two forces. 


Conditions to apply the Lami’s theorem 

There are following four conditions, as mentioned below, in order to apply the Lami’s theorem.
We must have to remember that all the three forces acting on a body must be in a same plane.  

We must have to remember that these three forces must be concurrent i.e. these three forces must be acting through a same point. 

We must have to remember that these three forces must be non-collinear i.e. these forces must not be acting in a same line of action. 

We must have to remember that these three coplanar, concurrent and non-collinear forces acting on a body must be in equilibrium i.e. there must not be any acceleration in the body upon the application of these three forces. We can also say that there must be zero net force over the body when these three forces will be applied. 

If these three forces are not following the above mentioned conditions, Lami’s theorem will not be applicable. 


Explanation of Lami’s theorem 

Let us assume that there is an object or a body of any given shape, as mentioned below, are exerted by three forces A, B and C. Angles made by these three forces with each other are displayed here in following figure as α, β and γ. 



According to Lami’s theorem statement, we will have following equation as mentioned here. 

A α Sin α, A = K Sin α
B α Sin β, A = K Sin β
C α Sin γ, A = K Sin γ 

Or we can also redefine it as 

K = A/ Sin α = B/ Sin β = C/ Sin γ 

Therefore, according to Lami’s theorem statement, we will have following equation which could be used in order to secure the magnitude of the unknown force in a case where object will be exerted by three coplanar, concurrent and non-collinear forces and object or body is in equilibrium.  


Limitations of Lami’s theorem 

There are few limitations of Lami’s theorem as mentioned here. 

There should be only three forces acting on the object. If there are more than three forces or less than three forces, we may not apply the Lami’s theorem. 

Lami’s theorem is applicable to only coplanar, concurrent and non-collinear forces. 

Therefore, we have seen here the basics of Lami’s theorem, proof of Lami's theorem and its limitations. 

Further we will find out another concept in engineering mechanics with the help of our next post.   

Do you have any suggestions? Please write in comment box and also drop your email id in the given mail box which is given at right hand side of page for further and continuous update from www.hkdivedi.com.     

Reference:  

Engineering Mechanics, By Prof K. Ramesh  
Image courtesy: Google    

Also read   

Thursday, 16 January 2020

January 16, 2020

APPARENT WEIGHT OF A MAN IN A LIFT

We were discussing the projectile motion - trajectory equation, definition and formulas and we have also seen the concept of Terminal velocity with the help of our previous posts.  

Now, we will be interested further to understand a very important topic in engineering mechanics i.e. apparent weight of a man in a lift. 

Today we will first understand the basics of apparent weight and true weight and further we will find out the apparent weight of a man in a lift for various situations. 

Apparent weight and true weight 

True weight of a body is basically defined as the force exerted by the earth on it. 

Let us think that a person whose mass is m and he is located at or near the earth surface, true weight or simply weight of the person will be mg. Where g is the acceleration due to gravity. 

Weight of the body will be measured with the help of weighing machine. When a body will be placed over the surface of weighing machine platform, there will be acting basically two types of forces.
  1. Force due to gravity of weight of the body i.e. W = mg 
  2. Upward normal contact force i.e. N exerted by the platform of weighing machine on the body 

APPARENT WEIGHT OF A MAN IN A LIFT

The upward normal contact force (N) exerted on the body will be termed as the apparent weight of the body. Value of the upward normal contact force will be dependent over the state of motion of the body. 

Apparent weight of a man in a lift 

Now let us see the apparent weight of a man in a lift for the various state of motion of the lift. Please note that we are going to use here the Newton’s law of motion in order to secure the expression for the apparent weight of a man in a lift for the various state of motion of the lift. 

I hope all of you are well aware with the lifts or elevators used in various multi-storey buildings or towers. If you have an opportunity to travel in a lift that travels longer distance such as lift of a 50 storey or 70 storey buildings, you will surely feel that there is acceleration, deceleration and steady state motion too. 

So, we will analyse and find out the apparent weight of a man for each state of motion of the lift and we will also see here a case of snap of lift or elevator with the help of this post. 

Case 1: Apparent weight of a man in a lift when lift is moving up and accelerating 

Let us consider that a man is standing inside a lift over the spring scale as displayed here in following figure. 

Let us think that lift is moving up with an acceleration a.  

APPARENT WEIGHT OF A MAN IN A LIFT

Let us draw the free body diagram, as displayed above, considering the man as a particle. I will determine the apparent weight of man by using the newton’s law of motion and please note that I am not standing inside the lift. I am standing outside of the lift on ground and analysing the apparent weight of man in the lift. 

∑ Fy = m x ay
R – mg = ma
R = m (g + a) 

As we have already seen here that apparent weight is basically the reaction force exerted on the body and hence apparent weight will be higher as compared to the actual weight of the man. 

Spring scale will display the weight of man equivalent to the reaction force exerted on the man by the surface of spring scale. 

Therefore, apparent weight of a man in a lift, going upward with an acceleration, will be higher than his actual weight or true weight. 

Case 2: Apparent weight of a man in a lift when lift is moving with a constant velocity i.e. acceleration is zero 

Let us consider that a man is standing inside a lift over the spring scale as displayed here in following figure. 

Let us think that lift is moving with a constant velocity V. We will see here the apparent weight of man and we will find out here that what will be the reading of spring scale.  


Let us draw the free body diagram, as displayed above, considering the man as a particle. I will determine the apparent weight of man by using the newton’s law of motion and please note that as lift is moving with a constant velocity and hence acceleration will be zero.   

∑ Fy = m x ay
R – mg = 0
R = mg   

As we have already seen here that apparent weight is basically the reaction force exerted on the body and hence apparent weight will be same as the actual weight of the man. 

Spring scale will display the weight of man equivalent to the reaction force exerted on the man by the surface of spring scale. 

Therefore, apparent weight of a man in a lift, going with a constant velocity, will be equal to his actual weight or true weight. 

Case 3: Apparent weight of a man in a lift when lift is moving down and accelerating 

Let us consider that a man is standing inside a lift over the spring scale as displayed here in following figure. 

Let us think that lift is moving down with an acceleration a.  


Let us draw the free body diagram, as displayed above, considering the man as a particle. I will determine the apparent weight of man by using the newton’s law of motion and please note that I am not standing inside the lift. I am standing outside of the lift on ground and analyzing the apparent weight of man in the lift. 

∑ Fy = m x ay
R – mg = - ma
R = m (g - a) 

As we have already seen here that apparent weight is basically the reaction force exerted on the body and hence apparent weight will be lower as compared to the actual weight of the man. 

Spring scale will display the weight of man equivalent to the reaction force exerted on the man by the surface of spring scale. 

Therefore, apparent weight of a man in a lift, going downward with an acceleration, will be lower than his actual weight or true weight. 

Case 4: Apparent weight of a man in a lift when lift cable is snapped 

Let us consider that a man is standing inside a lift over the spring scale as displayed here in following figure. 

Let us think that lift cable is snapped and lift is free falling. Lift will free fall with an acceleration g i.e. acceleration due to gravity.  

APPARENT WEIGHT OF A MAN IN A LIFT

Let us draw the free body diagram, as displayed above, considering the man as a particle. I will determine the apparent weight of man by using the newton’s law of motion and please note that I am not standing inside the lift. I am standing outside of the lift on ground and analyzing the apparent weight of man in the lift. 

∑ Fy = m x ay
R – mg = - mg
R = 0 

As we have already seen here that apparent weight is basically the reaction force exerted on the body and hence apparent weight will be zero.   

Spring scale will display zero as the apparent weight of man. 

Therefore, we have seen here the basics of apparent weight and true weight of a body with the help of this post. We have also secured here the apparent weight of a man in a lift for various situations. 

Further we will find out another concept in engineering mechanics i.e. Lami's theorem with the help of our next post.   

Do you have any suggestions? Please write in comment box and also drop your email id in the given mail box which is given at right hand side of page for further and continuous update from www.hkdivedi.com.   

We will find out now the Lami's theorem in our next post.   

Reference:  

Engineering Mechanics, By Prof K. Ramesh  
Image courtesy: Google    

Also read   

Tuesday, 14 January 2020

January 14, 2020

TERMINAL VELOCITY AND ITS EXPRESSION

We were discussing the projectile motion - trajectory equation, definition and formulas with the help of previous post.  

Now, we will be interested further to understand a very important topic in engineering mechanics i.e. Terminal velocity with the help of this post. We will find out here the basics of terminal velocity and we will also find out the expression for terminal velocity here in this post. 

So, what is terminal velocity? 

When an object starts to fall from initial zero velocity, object will move towards downward direction. Velocity of object will be increasing as the object will move towards downward direction under the constant acceleration i.e. acceleration due to gravity. 

Object will be subjected here to two external forces. First one is the weight of the object and second one is the drag force due to the air resistance. 

Now we will understand here the effect of air resistance on the object which is falling through the atmosphere. Drag force will be developed due to the effect of air resistance. 

Following expression, as mentioned below, provides the drag force due to air resistance.  

Where,
Fd = Drag force
Cd = Drag co-efficient
A = Projected Area
V= Velocity 

We can conclude that the drag force will be increasing with increase in the velocity of the object as drag force is directly proportional to the square of the velocity as we can see from the expression of drag force. 

There will be a situation where this drag force will become equal to the weight of the object and object will come under equilibrium and there will be no net force over the object and the vertical acceleration will go to zero.  


According to Newton’s first laws of motion, object will fall with constant velocity when drag force will become equal to the weight of the object and object will come under equilibrium and there will be no net force over the object and the vertical acceleration will go to zero. 

This constant vertical velocity will be termed as terminal velocity and it will be given by the following expression. 

Following table, as displayed below, indicates the terminal velocity for various cases and also indicates the distance needed to secure the 95% of the terminal velocity.  


Therefore, we have seen here the basics of terminal velocity, expression of terminal velocity, and significance of drag force with the help of this post. We have also secured here the terminal velocity for various cases and also indicates the distance needed to secure the 95% of the terminal velocity. 

Further we will find out another concept in engineering mechanics i.e. apparent weight of a man in a lift with the help of our next post.   

Do you have any suggestions? Please write in comment box and also drop your email id in the given mail box which is given at right hand side of page for further and continuous update from www.hkdivedi.com.  

We will find out now the apparent weight of a man in a lift in our next post.   

Reference:  

Engineering Mechanics, By Prof K. Ramesh  
Image courtesy: Google   

Also read  

Saturday, 11 January 2020

January 11, 2020

PROJECTILE MOTION - TRAJECTORY EQUATION, DEFINITION AND FORMULAS

We were discussing the importance of friction i.e. positive and negative effects of frictionClassification of friction and Coulomb's law of dry friction with the help of our previous post.  

Now, we will be interested further to understand a very important topic in engineering mechanics i.e. projectile motion - trajectory equation, definition and formulas with the help of this post. We will find out here the basics of projectile motion, classification of projectile motion and various equations and formulas associated with the projectile motion. 

Let us first start here with the basic definition of projectile motion 

Projectile motion is basically defined as a motion where a particle moves in a vertical plane with some initial velocity but its acceleration will always be the free fall acceleration i.e. acceleration due to gravity which will be acting towards downward direction. 

When a particle will be thrown in to the space, it will have motion in x direction i.e. in horizontal direction and it will have motion in y direction too i.e. in vertical direction. 

The combination of motion of particle in x direction i.e. in horizontal direction and in y direction i.e. in vertical direction could be considered as the projectile motion. 

We must need to note it here that the horizontal motion and vertical motion of the particle in a projectile motion will be independent with each other. 

Let us consider that a particle is thrown in to the space with initial velocity u with an angle θ with the horizontal direction as displayed here in following figure. 


Following equations, as displayed below, will be used in order to determine the various desired formulas for a particle following the projectile trajectory. 


Where,
u = Initial velocity of the particle
V = Final velocity of the particle
a = Acceleration of the particle
t = Time of travel of the particle    

Above equations will be only valid for a particle which is under motion with constant acceleration.
In case of projectile motion, a i.e. acceleration of the particle will be basically acceleration due to gravity i.e. g and it will be acting towards downward direction. 

We will have following equations, as mentioned below, for particle motion in horizontal and in vertical direction. 


Projectile trajectory equation 

Above equation shows that it will be a parabolic equation and hence projectile motion trajectory will be parabolic. 

Time taken to reach the ground 

Time taken to reach the ground i.e. T will be determined with the help of following equation as mentioned below. 

Range or horizontal distance traveled by the particle

Range or horizontal distance traveled by the particle will be determined with the help of following equation as mentioned below. 

Maximum height or maximum vertical distance traveled by the particle

Maximum height or maximum vertical distance traveled by the particle will be determined with the help of following equation as mentioned below. 


Classification of projectile motion

Projectile motion will be basically classified in two types as mentioned here.
  1. Horizontal plane projectile motion
  2. Inclined plane projectile motion 

If we have a problem or a case of projectile motion where the point of projection and point of striking both are on inclined plane, we will say such projectile motion as inclined plane projectile motion.
Otherwise, we will say horizontal plane projectile motion. 

We must note it here very clearly that for inclined plane projectile motion, the point of projection and point of striking both must be on inclined plane. 

Therefore, we have studied here the basics of projectile motion, classification of projectile motion, equations associated with projectile motion, time of flight, range and maximum height traveled by the particle in a projectile motion with the help of this post. 

Further we will find out another concept in engineering mechanics i.e. terminal velocity and its expression with the help of our next post.  

Do you have any suggestions? Please write in comment box and also drop your email id in the given mail box which is given at right hand side of page for further and continuous update from www.hkdivedi.com.  

We will find out now the terminal velocity and its expression in our next post.  

Reference:  

Engineering Mechanics, By Prof K. Ramesh  
Image courtesy: Google    

Also read