Most imp ques 5marks... Engg Mechanics for 2nd sem

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[00:00:01]Very good evening, my dear second semester diploma students.

[00:00:07]Now, I'm giving four important questions with answers. Subject is theory four, engineering mechanics. Now, I am discussing one by one.

[00:00:22]Important question. Two men carry a weight of 2 kN by means of two ropes fixed to the weight.

[00:00:29]One rope is inclined at 45° and the other at 30° with their vertices.

[00:00:35]Find the tensions in the each rope.

[00:00:38]Here's the diagram.

[00:00:40]A B C. This angle is 30, this angle is 45.

[00:00:43]This load is 2 kN.

[00:00:46]Now, from the free body diagram, T TBC, tensile force along the line BC.

[00:00:55]And B this B, then uh TAC, TAC meaning tensile force along the line AC. That is AC is the BC.

[00:01:07]Now, from the free body diagram and according to Lami's theorem, TBC by sin 90 + 45 because this is the tensile force TBC, then opposite angle is uh sin 90 + 45.

[00:01:25]Then, tensile force TC, the opposite angle is tan ni- tan sin 90 + 60.

[00:01:34]Then, uh the force two, the opposite angle is sin 75 because total angle is 180, this angle is 60, this angle is the 45. Therefore, 180 - 60 + 45 is equal to 180 105 is equal to 75.

[00:02:00]Therefore, according to Lami's theorem, TBC by sin 90 + 45 is equal to TSC by sin 90 + 60 is equal to 2 by sin 75.

[00:02:10]Or TBC by sin 135 TSC by sin 150 is equal to 2 by sin 75.

[00:02:17]Then, TBC the from the calculator, we have sin 135 is equal to 0.7071.

[00:02:26]Sin 150 is equal to 0.5. TSC is equal to 0.5.

[00:02:30]Is equal to 2 by 0.9659 because sin 75 is equal to 0.96 59.

[00:02:38]Then, tensile force along the line BC by 0.7071 is equal to 2 by 0.9659.

[00:02:48]Therefore, tensile force along the line BC is equal to 2 into 0.7071 by 0.9659 is equal to 1.46 kN.

[00:03:00]Then, tensile force along the line AC, TSC by 0.5 is equal to 2 by 0.9659.

[00:03:09]Or TSC is equal to 1.035 kN.

[00:03:14]From the above calculation, we find that the tension at AC rope is 1.035 kN and tensions at BC rope 1.46 kN. It is the most important question. Then, another question. Derive the expression for maximum mechanical advantage.

[00:03:36]Uh you know that mechanical advantage is equal to W by P.

[00:03:40]A law of machine is given by P is equal to MW plus C.

[00:03:44]Where P is equal to effort applied, W is equal to load lifted, M is equal to slope of the line, C is equal to Y intercept of the straight line.

[00:03:53]Then MA is equal to W by P is equal to W by MW plus C.

[00:03:59]Is equal to W by W by MW by W plus C by W because dividing both numerator and denominator by W.

[00:04:08]Therefore, we have MA is equal to 1 by M plus C by W.

[00:04:12]As C is less than W, the ratio C by W is very small and can be neglected. Therefore, mechanical advantage maximum is equal to 1 by M.

[00:04:26]And the maximum efficiency is equal to MA by MA maximum by VR.

[00:04:31]Then you know that MA maximum is equal to 1 by M by VR.

[00:04:36]Therefore, maximum efficiency is equal to 1 by M into VR.

[00:04:44]Now, another question. A screw jack has a thread of pitch equal to 10 mm.

[00:04:50]Find the effort to be applied at the end of the handle 500 mm long to lift a load of 1800 N.

[00:04:58]The efficiency of the system is 52%.

[00:05:02]First data given, pitch small P is equal to 10 mm.

[00:05:06]L 500 mm.

[00:05:08]W load is equal to 1800 N. Efficiency is equal to 52% is equal to 0.52 because 52 by 100 is equal to 0.52.

[00:05:19]P is equal to effort required.

[00:05:21]You know that velocity ratio is equal to 2 pi L by P.

[00:05:28]Velocity ratio for screw jack 2 pi L by P.

[00:05:32]Then 2 pi into 500 by 10 is equal to 314.16.

[00:05:39]Then mechanical advantage W by P 1800 by P or 0.52 because efficiency is equal to 0.52.

[00:05:49]It is equal to 1800 by P by 314.16.

[00:05:54]After calculating, we have P is equal to 11.018 N.

[00:05:59]Therefore, effort required is 11.018 N.

[00:06:04]Then, another question is similar type of questions.

[00:06:08]A screw jack has a thread of 10 mm pitch.

[00:06:11]Where what effort applied at the end of a handle 400 mm long will be required to lift a load of 2 kN if the efficiency at this load is 45%.

[00:06:23]First pitch, small P.

[00:06:26]P is equal to pitch of thread 10 mm.

[00:06:28]Length of handle 400 mm.

[00:06:31]Efficiency is equal to 45% is equal to 0.45.

[00:06:35]P is equal to effort required to lift the load. Uh you know that velocity ratio is equal to distance traveled by the effort by distance traveled by the load is equal to 2 pi L by P.

[00:06:48]After putting all the values, we have velocity ratio is equal to 251.42.

[00:06:54]Again, mechanical advantage is equal to W by P.

[00:06:58]It is equal to 2 kN by P is equal to 2 into 1000 by P because 1 kilo 1 kilo is equal to 1000.

[00:07:05]2000 by P.

[00:07:07]We also know that efficiency is equal to MA by VR.

[00:07:11]After putting all the values, efficiency 0.45 is equal to 2000 by P divided by 251.42.

[00:07:22]Then, 0.45 is equal to 2000 by P into 251.42.

[00:07:28]After calculating, we have P is equal to 17.67 or it is equivalent to 17.7 N.

[00:07:37]P here problem practice for the year.

[00:07:41]Mr. I successful over. Best of luck for your examinations.

[00:07:45]Mr. successful however best of luck for your examinations.