Mechanical Engineering: Ch 11: Friction (34 of 47) Belt Friction: Ex. 3
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- Опубліковано 27 сер 2024
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In this video I will find the moment on the larger drum of a belt drive around 2 drums.
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sorry sir why don't you use (800-384)*0.15*(COS20)? Doesn't the tension have a x and y component as well?
Then tension in the belt will always be in the direction of the belt. (no components).
sir what will be B(Beta) angle in case of pulley B?
Nice bow tie 👌 you should explain why your analysis is all based on the smaller pulley ! Hint : draw the motor !
GREAT content! I have a question? Can a pulley system enable a diesel engine to achieve a greater RPM than its rating. for instance, i have a Detroit diesel engine that's gross rated maximum power is 300 bhp @ 2800 RPM. that is powering a positive displacement blower that's maximum rating is 3000 RPM @ 135 bhp. In order to maximize the output of the blower, I need to increase the RPM at the pump to 3000 RPM. that's 200 RPM above the engines gross rated maximum RPM of 2800, would a pulley system allow me to do this?
Engines are usually designed to provide a certain torque, (besides raw horse power), and they will be more or less efficient at different torque requirements. A pulley will change the torque, so you'll have to take that into account.
Sir, I have a textbook question that I attempted. Sand drops vertically at the rate of 2kg\s on to a conveyor belt moving horizontally with a velocity of 0.1m\s. Calculate (i) the extra power needed to keep the belt moving, (ii) the rate of change of kinetic energy of the sand. The answer (i) 0.02W (ii) 0.01W. Why are these two different?
Yes, it seems very odd, but the book is correct. The change in KE = (1/2) m (change in v)^2 = 0.01 W But to find the power required we need to consider the change in momentum: W = F x d P = W / t Impulse = I = F x t = change in momentum = m x (change in v). Therefore P = W / t = F x d / t = F x v = m x (change in v) x v / t = 0.02 W
good job
Thanks Sir I understood . Sir part (iii) of that question asks why the extra power needed is twice as great as the rate of change of kinetic energy?
Sir could you explain.
Thank you.
Are these forces difference that leads to the vibrations of the rope in real life?
Vibrations have specific causes which are not related to the tension, (however tension will play a role).
What would you do if the angle of contact not given?
Thanks Sir I understood but what happened to the the difference of 0.02W-0.01W ?
The 2 results are independent. The power is needed to change the momentum of the sand. The increase in kinetic energy has nothing to do with the change in momentum. (Note that in collisions, the momentum is always conserved and energy is never conserved)
youre the best. Thanks!
Sir where you get the Tmax which is 800N? Thanks and God bless 😇
I believe that is a given in the problem?
please sir, can you please upload videos for 1st order circuits, 2nd order circuits, phasor domain, power of sinusoidal signals, bode plots, convolution and two port circuits for Electrical engineering because this would help me alot to pass my 2nd module in my bachelor course. thank you
how come we dont consider the friction coefficient between the large drum and the belt? wouldnt the moment applied be affected? especially if the tension in the belt exceeds the static friction force?
Because we are only considering the effect on the right side. The assumption is that A is driving wheel B.
Why u need to multiply it by pi/180
To convert from degrees to radians
i get lost in 180 - 2(20)
The angle is needed to calculate what portion of the small wheel makes contact with the drive belt.
given that small angle sir, how can I know the angle that the belt makes contact to the wheel?
like, I forgot the geometry about it😅