Lecture 16 : Tutorial - 1 (Part 2 of 2)

[Music] [Music] welcome to the second part of the tutorial session of the content test was covered in week one we had discussed about units and dimensions and scales and vectors in the previous part of the tutorial here I'll discuss about the Statics skinetics and kinematics and laws of motion some problem statements would be solved just to recall the basic Theory or in Statics equilibrium of forces sum of all the forces acting on a body must be zero for the body to be in a static equilibrium that means summation of all the forces should be equal to zero equilibrium of Ms some of all movements about any point must be zero for the body to be in static equilibrium so that means sum of all the movements is also equal to zero moment of a force the moment M of a force F about a point is product of the force and the perpendicular distance R from the point to the line of action of the force that means moment is equal to for Force into R these are certain principles that we discussed in the week one simple relationships for the velocity which is change in the position of the object that is displacement per unit change in time this is velocity velocity of rate of change of displacement with respect to time similarly acceleration a isal to DV by DT that is change in velocity per unit change in time it is rate of change of velocity with respect to time then the equations of motion that we discussed first equation of motion that is V that is final velocity is equal to U + a final velocity V is the sum of initial velocity and the product of the acceleration and time second equation of motion is s isal to U +/ a² that is displacement s is the sum of of the initial velocity into time plus half of the product of acceleration and the square of the time third equation of motion that is U v² = to U ² + 2 a s that is square of the final velocity is equal to the square of the initial velocity plus twice the product of acceleration and displacement these are General formulas that you had been reading in your senior secondary education as well and these would be applied in engineering Mechanics for then we had Newton Second Law that is force is equal to mass into acceleration that is force acting on an object is equal to the mass of the object multiplied by its acceleration work energy principle that is work is equal to Delta K that is work done in an object is equal to change in the kinetic energy then power power is equal to change in the work per unit change in time it is rate at which work is done over time now let us try to use these to solve some numerical questions a 10 kg block is an equilibrium on a frictionless surface with forces fub1 isal to 20 nton and FS2 find the magnitude and direction of F2 the body stays in equilibrium that is the first principle of equilibrium that we saw that means suppose this is a body frictional surface is there it is a 10 kg block F1 force is 20 nton which will try to push this in this direction to keep it in equilibrium we have to find what will the value of FS2 simply we know summation of all the force is equal to zero that means fub1 + FS2 is equal 0 fub1 is 20 nton + F2 is = 0 that means FS2 is = - 20 Newton that is in the opposite direction that means I have my Force FS2 here if this is my fub1 is equal to 20 Newton FS2 would be equal to minus of 20 Newton right this is a simple application of the equilibrium of forces principle to find the force acting in the opposite direction so there's a problem statement where it is mentioned a car with a mass of 500 kg moving with a velocity of 10 m/s calculate the kinetic energy we all know there is a simple relation between kinetic energy and the mass and velocity mass given is 500 kg and velocity given is 10 m/s by slow moving car as know that is given kinetic energy relationship we know is equal to half M v² which is equal to half into Mass into square of the velocity everything is given in SI units me/ second 10 square this is equal to 25,000 jewles so this is very simple problem statement I'll give you another problem statement where you have a velocity of a tennis ball it is moving at 40 m/ second and kinetic energy is given 80 Jew calculate the mass using the same relationship kinetic energy is equal to half M Square you need to find the mass of the tennis ball this problem you solve by yourself you'll find the solution in the lecture notes another problem statement is on the rotation now the angular velocity a dis rotates with an angular acceleration of 2 radians per second if the initial angular velocity is 3 radians per second find the angular velocity after 4 seconds and angular displacement during this time this is interesting so we have a mega KN that is the initial angular velocity is 3 radians per second angle acceleration which is dened by Alpha is equal to 2 radians per second squar and we have been given time is equal to 4 seconds so angular velocity is equal to Omega is equal to we use the similar relationship like V = to U + a this is omega = omega plus Alpha into T simply we have omega note as 3 plus Alpha is 2 into time is 4 which is equal to 11 radians per second angular velocity right then I have angular displacement Theta this equal to Omega t+ half Alpha t² this is = 3 into 4 + 1 by 2 into Alpha is 2 t² is 4 S this gives me 28 radians that is angular displacement during this time is 28 radians there's another problem statement a box is sliding down a friction L incline of 30° if the box starts from rest find the velocity after 5 seconds so here incline angle Theta is equal to 30° and uh the box starts from rest so here we know acceleration because the angle given is 30° this is a plain surface 30° and this is a box that is sliding down this frictionless surface so acceleration is equal to a = g sin Theta this equal to 9.81 into sin of 30 this is = to 9.81 into 1X 2 is equal to around 4.90 5 m per second squared initial velocity U equal to 0 final velocity is to be found it is v = u + a t this is equal to initial velocity is zero acceleration is 4.95 into Time 5 this is equal to 24 5 to 5 m/s the velocity after 5 seconds is 24.5 5 m/s so there are certain other problems that you can solve by yourself next topic is Laws of Motion inertia and momentum certain relations which were discussed the first law of inertia that the body at rest or in uniform motion will remain so unless acted upon by an external force that is a force is equal to zero that is V is constant second law that we've already taken in the previous slide as well the acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass is force into Mass into acceleration Third Law of the motion given by Newton is of the equalent opposite reaction that is for every action there is an equalent opposite reaction that is fub1 2 is equal to minus of F21 then we talked about the linear momentum it is equal to P is equ Al to mass into velocity the momentum units is kg into m/s Mass into velocity impulse is J that is change in momentum which is also equal to force into change in time so there's a theorem known as impulse momentum theorem that is force per unit change in time is equal to mass per unit change in velocity conservation of momentum that is initial momentum and final momentum is same in a closed system with no external forces inertia moment of inertia rotational analog of mass of linear motion we know I equal to Mr r² units are kg into me squ rotational inertia to is equal to I into Alpha that is talk is equal to the moment of inertia times the angular acceleration let us see a few numerical problems here find out how much net force will be needed to accelerate a 2500 kg truck at 5.5 m/s squar so we have to accelerate it to this level so we know here acceleration a is equal to 5.5 m/s square and mass is given that is 25 100 kg we need to find the force force is simple Mass into acceleration that is 2500 into a 5.5 that is equal to 13750 Newton so net force required or needed to acceleration a 2500 kg truck to an acceleration of 5.5 at an acceleration of 5.5 m/s squar is 13750 Newton let me try to see another example where a racing car driver is on a RAC trck in order to overtake he accelerates his racing car first at the rate of 8 m/s square then at the rate of 16 m/s squ find the ratio of the forces exerted by the engine for the acceleration change so we need to find the ratio of the forces which it has exerted so it is simple ratio to be taken so we will just use a simple formula again force is equal to mass into acceleration that is force fub1 is equal to mass 1 into acceleration 1 and Force 2 is equal to Mass 2 into acceleration 2 the mass is same all the time a mass is M so we know A1 the initial acceleration here given is 8 m/s square and we know A2 is equal to 16 m/s square so we have to find the ratio between fub1 and FS2 fubb by F2 is equal to M A1 by m A2 that is equal to A1 by A2 simply we have A1 by A2 is = 8 by 16 is equal to 1 by 2 that means fub1 ratio F2 is 1 ratio 2 so ratio of the forces is exerted by the engine is 1x two for that is we have taken or gotten result here there's another example that is here when a bullet of mass 20 G is shot from a gun that has an initial velocity of 40 m/ second the mass of the gun is 5 kg what is the initial recoil velocity of the gun so it has been given the mass of the bullet as 20 G we have to convert this into kg and we have been given the mass of the gun that is 5 kg right and initial velocity is given for the bullet is 40 m/s recall velocity of the gun we have to takeen so we will use use the law of the conservation of momentum and try to solve this problem you can solve it by yourself and try to find what is the initial recoil velocity of the gun try to solve it by yourself try to use the law of the conservation of momentum and get the solution you will get the solution in the lecture notes there you will find it and this is for you as an home assignment another law of conservation of momentum problem here is there are cars with masses two 2 kg and 5 kg respectively the toy cars that are at rest a car having Mass 5 kg moves towards the east with a velocity of 5 m/s find the velocity of the car with mass 2 kg with respect to ground so we have M1 as 2 kg and M2 as 5 kg so V1 is to be determined and V2 yeah it is given 5 m/s V2 is equal 5 m/s so from the law of conservation momentum we know that the final momentum is equal to M1 V1 + M2 V2 so that is 2 kgs into V1 + 5 kgs into V2 that is 5 m/ second also we know that initi momentum is zero because they were at rest here it is given the cars were at rest we also know initial momentum is equal to final momentum Law of Conservation momentum once again so therefore this 0 isal = to 2 kg into V1 plus 25 kg into m/s this gives me the value of V1 as 12.5 m/s this tutorial session had very trival problems trival problems means the problems that you have been studying in the first year engineering or maybe in your pre-engineering senior secondary education in the fourth coming lectures where we will talk about the tutorial sessions on this Tess train diagram about the different mechanical components clutches brakes belts Etc that will be having application of all these laws that we have studied here so I will give you a sheet of the numerical problems and statements that you'll also solve for your home assignment with this I'm ending this lecture thank you [Music] [Music] [Music]