Recent content by JimmyTheBlue

Doing the same question now alas, for part B did you calculate the gearbox torque magnitude via the assumption that it dissipates 8% of the initial power? Or just via the assumption that T(Driver) = T(Acceleration) + T(Frictional) which semi amounts to the same thing.

The system isn't under acceleration, so we can say the input and output powers are equal? If that's the case then... Driver speed = 250 revs min-1 = 25pi/3 rad s-1 Driven Speed = (25pi/3) * (0.075/0.2) = 25pi/8 rad s-1 Power dissipated at pulley 2 = (25pi/8) * 200 = 625pi W (This is the answer?)...

Setting T1 to the maximum tensile strength seemed to be the books method of finding the maximum power input, but as you say that method pretty much calculates it without taking into account the load. We can find the speed of the second pulley assuming no slippage, and thus the power dissipated...

Was just looking at this one. A quick bit scrap programming on Python gave me - Diameter: 0.3 N1: 80 N2: 75 N3: 25 N4: 20 Speed: 400.0 Diameter: 0.3 N1: 80 N2: 80 N3: 20 N4: 20 Speed: 400.0 Mathematically - -> We know N1 = 4N4 by the gearing ratio alone. -> We know 300 = N1 + 2N2 + 2N2 + N4 ->...

Homework Statement A pulley 150 mm diameter is driven directly by an electric motor at 250 revs min-1. A V-belt is used to transmit power from this pulley to a second pulley 400 mm diameter against a load of 200 Nm. The distance between the centre of the pulleys is 600 mm, the included...

He calculated the radius via the volume and the height. He converted the yield strength to PA then calculated the hoop stress from that using the safety factor. From that he used Hoop stress = pr/t to find the thickness. The rest is self explanatory. He used the thin cylinder equations as I did...

Dammit. I totally didn't read that and regurgitated the formula in the book despite even spotting the half circle capacitor intricacy lol. Cheers. Recalc time.

"Presumably you mean (2n - 1) there, where n is the number of plates. In this case (2n - 1) = 7." I don't think that is the case. 4 plates would equate to 3 individual parallel capacitors which is where the multiplying factor of n-1 was tieing in.