Important for Mechanics💐📚

Added: 2026-05-06

[00:00:01]Very good evening. My dear second semester diploma students, today I am discussing um engineering mechanics two, five, and 10 marks type of questions. These questions are very important for your semester examinations.

[00:00:20]First, I am discussing one by one.

[00:00:26]First, possible questions for second semester examinations, engineering mechanics.

[00:00:32]Today, I'm discussing five marks type of questions. First question, define MA, VR, and efficiency of lifting machine.

[00:00:40]Derive the relation between them. You know that mechanical advantage the in the ratio of the load lifted to the effort applied is called mechanical advantage.

[00:00:51]Mathematically, MA is equal to W by P, where W is equal to load lifted, P is equal to effort applied. It always be your pure number.

[00:01:01]Then, velocity ratio.

[00:01:03]It is a ratio between distance moved by the effort applied and distance moved by the load lifted.

[00:01:10]It can be written as VR is equal to Y by X.

[00:01:16]Where VR is equal to distance moved by the effort applied Y divided by distance moved by the load lifted X. VR is equal to Y by X.

[00:01:27]Then, what is efficiency of lifting machine?

[00:01:30]It is the ratio of the useful work done by the machine to the work done on the machine. It can be expressed as efficiency is equal to output by input into 100.

[00:01:42]Then, what is the relationship between MA, VR, and efficiency of a machine?

[00:01:48]You know that efficiency this is a symbol for efficiency. Output by input is equal to W into X by P into Y.

[00:01:57]Let us consider a lifting machine. W is equal to load lifted by the machine.

[00:02:02]P is equal to effort required to lift the load.

[00:02:05]Where Y is equal to distance moved by the effort in lifting the load.

[00:02:10]X is equal to distance moved by the load lifted. You know that MA is equal to W by P, VR is equal to Y by X.

[00:02:20]Therefore, you also know that input of a machine P into Y.

[00:02:24]Output of a machine W into X.

[00:02:27]Then, efficiency is equal to output by input. W into X by P into Y is equal to W by P into X by Y.

[00:02:36]Or we can write as W by P into one by Y by X because X by X by Y is same equal to Y by X.

[00:02:46]Therefore, W by P by Y by X. You know that W by P is equal to mechanical advantage. Y by X is equal to velocity ratio.

[00:02:56]Therefore, efficiency is equal to MA by VR.

[00:02:59]Proved. This equation very important for your semester. Then, state the laws of friction.

[00:03:05]Actually, there are two types of laws of friction. Laws of static friction and laws of dynamic friction.

[00:03:12]The force of friction always acts in a direction opposite to opposite to the direction of motion between two surface.

[00:03:19]Force of friction acts along the common plane of contact between two bodies.

[00:03:25]The force of friction is equal to the external force applied for motions.

[00:03:32]Force of friction depends upon the roughness of the surface.

[00:03:36]Force of friction is independent of area contact.

[00:03:40]Laws of dynamic friction.

[00:03:42]The force of friction always acts in a direction opposite to the direction of motion of the body.

[00:03:48]The magnitude of dynamic friction bears a constant ratio to normal reaction between the two surface.

[00:03:55]The force of friction decreases with the increase of motion.

[00:03:59]Then, third. The greatest and the least resultant of two forces are respectively 17 kN and 3 kN.

[00:04:08]Determine the angle between the two forces when the resultant is root over 149 kW kN, sorry.

[00:04:18]Let P and the Q be the two forces under considerations.

[00:04:22]According to question, greatest resultant means P plus Q is equal to 17.

[00:04:28]Least resultant means P minus Q is equal to three. This is the equation one. This is the equation two.

[00:04:34]Now, adding equation one and the two, we have P plus Q is equal to 17, P minus Q is equal to three.

[00:04:41]Then, uh plus Q minus Q cancel. P plus P two P is equal to 20. Or P is equal to 20 by two, 10.

[00:04:50]Putting the value of P in equation one, we have P plus Q is equal to 17. Or 10 plus Q is equal to 17 because P is equal to 10.

[00:05:00]Or Q is equal to 17 minus 10 is equal to seven.

[00:05:04]You know that resultant of two forces P and the Q is R is equal to P squared plus Q squared plus two PQ cos theta root over.

[00:05:14]But, the value of R is given 149 root over. Then, after putting all the values, we have root over 149 is equal to 10 squared plus seven squared plus two into 10 into seven into cos theta root over.

[00:05:30]Uh square square of 10 is equal to 100.

[00:05:33]Square of seven is 49.

[00:05:35]Two into 10 into seven, 140 cos theta root over.

[00:05:40]Uh 149 plus 140 cos theta root over.

[00:05:43]Squaring both sides, we have 149 root over squared is equal to 149 plus 140 cos theta squared uh root over squared.

[00:05:53]Then, we have 149 is equal to 149 plus 140 cos theta.

[00:06:01]Or 140 cos theta is equal to 149 minus 149, zero.

[00:06:05]Or cos theta is equal to zero by 140 is equal to zero. Or theta is equal to cos inverse zero uh is equal to 90 degree. Or we can write as cos theta is equal to cos 90 because since cos 90 is equal to zero.

[00:06:21]Therefore, theta is equal to 90 degree.

[00:06:24]Then, number four. Define beam and explain different types of beam.

[00:06:29]What is beam? Beam is a structural member which is acted upon by a system of external loads at right angles to the axis.

[00:06:38]Now, I'm discussing different types of beam.

[00:06:42]First, cantilever beam. A cantilever beam is one which is fixed at the one end. It is a fixed at the one end and a free at the other end. In this figure, uh end A is rigidly fixed into its supports and the other end B is free.

[00:07:00]Examples are Tata.

[00:07:02]The length between A and B is known as length of cantilever.

[00:07:08]Then, what is simply supported beam?

[00:07:11]A simply supported beam is one whose ends freely rest on walls or columns.

[00:07:18]In all such cases, the reactions are always upwards. In these cases, all all reactions are always upwards.

[00:07:26]Then, overhanging beam. An overhanging beam is one in which the supports are not situated at the ends in one or both ends.

[00:07:36]Here, supports C and the D not at ends. In figure, C and D are two supports and both the ends A and the B A and the B of the beam are overhanging beyond the supports C and the D respectively. This is the diagram of a overhanging beam.

[00:08:00]What is fixed beam? A fixed beam is one whose both ends are rigidly fixed or built in to supporting walls or column.

[00:08:09]This is the fixed beam.

[00:08:13]Now, what is continuous beam? A continuous beam is the beam on which it has more than two supports.

[00:08:22]Because one, two, three, four are more than two supports. These supports are the extreme left and right are called the end supports and all the other supports except the extreme are called intermediate supports. These supports are called intermediate supports.

[00:08:42]Then, these important questions very important. Uh in a lifting machine, an effort of 30 N just just lift a load of 720 N. Find the mechanical advantage.

[00:08:55]If efficiency of machine is 30%, at the loads, calculate the velocity ratio of the machine.

[00:09:01]You know that uh P is equal to effort 30 N, W is equal to load lifted 720 N, efficiency is equal to 30% is equal to 0.3.

[00:09:11]You know that mechanical advantage is equal to W by P. That is load lifted by effort applied. Here, W is equal to 720, P is equal to 30 is equal to 24.

[00:09:23]You know that efficiency is equal to MA by VR.

[00:09:27]Or 0.30 efficiency 0.30 given.

[00:09:32]24 by velocity ratio. Or velocity ratio into 0.30 is equal to 24.

[00:09:39]Or velocity ratio is equal to 24 by 0.30 80.

[00:09:45]Then, number six. An I-section has the following dimension. Bottom flange.

[00:09:51]This is the bottom flange. Dimension is 400 into 100.

[00:09:55]Web. This is called oil. Oil 400 into 100. This length is 400. This width is 100.

[00:10:03]Then top flange 200 into 50. This length is 200. This width is 50.

[00:10:12]Calculate.

[00:10:16]Determine the center of the given section.

[00:10:20]Now, the I-section is symmetrical about YY axis.

[00:10:25]Why it is called YY axis? Suppose But this distance is equal to this distance. This distance is equal to this distance. This distance is equal to this distance. Therefore, this I-section is symmetrical about YY axis. And bottom flange of the be the axis of the I-section.

[00:10:50]Now, bottom flange A1 means area. 400 into 100. Because this length is 400.

[00:10:58]This length is 100. Therefore, area is equal to 400 into 100.

[00:11:04]is equal to 40,000 mm squared. Y1 Y1, this length is 100. Therefore, 100 by 2 is equal to 50.

[00:11:14]>> [clears throat] >> 50 Oil, A2 is equal to 400 into 100.

[00:11:19]40,000 mm squared.

[00:11:21]Y2 is equal to 100 plus 400 by 2 is equal to 300.

[00:11:31]2 This is 1. This is 2. This is 3.

[00:11:35]Y1 100 by 2 constant 100 plus This height is 400. 400 by 2 is equal to 300.

[00:11:48]Therefore, 100 plus 400 by 2 is equal to 300.

[00:11:52]Top flange A3 is equal to 200 into 50.

[00:11:56]This length is 200. This length is 50.

[00:11:59]Therefore, 200 into 50 10,000 mm squared. Y3 Y3 A-section length This height is 100. This height is 400 plus this height is 50. Therefore, 50 by 2.

[00:12:14]Therefore, Y3 is equal to 100 plus 400 plus 50 by 2.

[00:12:21]>> [snorts] >> 100 plus 400 plus 50 by 2 is equal to 525 mm squared. Then Ybar, you know that Ybar is equal to A1Y1 plus A2Y2 plus A3Y3 divided by A1 plus A2 plus A3.

[00:12:37]>> [snorts] >> Then after putting all the values, Ybar is equal to 40,000 into 50 40,000 into 300 10,000 into 525 divided by 40,000 plus 40,000 plus 10,000.

[00:12:54]Then after calculating When I am going to run the program Then Ybar is equal to 213 plus 880 mm.

[00:13:08]Then another out two questions are important.

[00:13:12]A body is acted upon by three forces. 2 kN 2 kN, 4 kN, and 1 kN.

[00:13:20]Force 2 kN is horizontal. This 2 kN is horizontal.

[00:13:26]And [snorts] 4 kN acts 45 degree degree to the horizontal. And inclined 45 degree.

[00:13:35]Then 1 kN 1 kN is vertical. Determine the resultant of three forces. You know that delta H algebraic sum of the horizontal forces.

[00:13:48]horizontal force cos vertical force sin Therefore, delta H is equal to 2 cos 0. This force is 2. Angle is 0. 2 cos 0 plus 4 cos This angle is 45. plus 1 cos 90 Then after calculating from calculator, we have 4.828 kN. Delta V algebraic sum of the vertical forces 2 sin 0 plus 4 sin 45 plus 1 sin 90 is equal to 3.828.

[00:14:24]R is equal to delta H squared plus delta V squared all root over is equal to 6.1614 kN. This is the resultant.

[00:14:35]Then What is theta? tan theta You know that tan theta is equal to delta V by delta H. 3.828 by 4.828 0.7392 theta is equal to tan inverse 0.792 is equal to 38.40.

[00:14:52]Then number seven >> [snorts] >> An electric light fixture of weight 20 This 20 N is supported as shown in the figure. Determine the forces in the wires AC and BC.

[00:15:08]Let T1 tensile force along the line AC T2 tensile force along the line BC >> [snorts] >> Here point C is in equilibrium under three forces T1, T2, and 20 N. This is the free body diagram of this fixture.

[00:15:29]>> [snorts] >> Now, here this angle is 45.

[00:15:32]This angle is also 45. This angle is 60 degree. This angle is 60 degree. But this angle is 75 because this total angle is 180. This angle 60.

[00:15:46]This angle 45. 60 plus 45 105 180 [snorts] minus 105 is equal to 75 degree.

[00:15:55]Then according to Lami's theorem T1 by sin 90 plus 45 is equal to T2 by sin 90 plus 60 20 by sin 75 Here T1 by sin 135 is equal to T2 by sin 150 is equal to 20 by sin 75 0.9659 After calculating, we have T1 is equal to 14.6412 N.

[00:16:20]T2 is equal to 10.35 N.

[00:16:23]>> [snorts] >> This T1 opposite angle sin [snorts] 90 plus 45 because this is 90. This is 45.

[00:16:34]T2 [snorts] opposite angle theta This angle is 90. This angle is 60.

[00:16:38]Therefore, sin 90 plus 60 >> [snorts] >> Then last question, the resultant of two equal forces acting at a point also equal to P. Determine the angle between the two forces.

[00:16:52]>> [snorts] >> In this figure, two forces are equal to P. And the resultant force is also equal to P.

[00:16:59]You know that R is equal to P squared plus Q squared plus 2PQ cos theta.

[00:17:04]But then according to question P is equal to P squared plus P squared plus 2P into P cos theta because P is equal to Q is equal to P.

[00:17:14]2P squared plus 2P squared cos theta all root over or P is equal to 2 Taking common 2P squared within bracket 1 plus cos theta or 2P squared plus within bracket 1 plus cos squared theta by 2 minus sin squared theta by 2 because cos theta is equal to cos squared theta by 2 minus sin squared theta by 2.

[00:17:35]Then or 2P squared cos squared theta by 2 minus plus cos squared theta by 2 because 1 minus sin squared theta by 2 is equal to cos squared theta by 2.

[00:17:48]>> [snorts] >> Therefore, P is equal to 2P cos theta by 2.

[00:17:53]or P is equal to 2P cos theta by 2 or P by 2P 2P 2P is equal to cos theta by 2. P by 2P is equal to 1/2. is equal to cos theta by 2. You know that cos 60 degree is equal to 1/2.

[00:18:09]Therefore, I am writing 1/2 cos 60 is equal to cos theta by 2. Then cos cos cancel.

[00:18:18]theta by 2 is equal to 60 degree. or theta is equal to 2 into 60 degree is equal to 120 degree.

[00:18:25]But then after the question practice for Then next video 2 minute 5 minute 10 minute video Then you can practice for Best of luck for your examinations.