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Pulley Torque calculations 
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#1
Oct1811, 04:04 PM

P: 6

I think I'm over thinking this, so someone set me straight.
If I have an eletric motor with a 75mm OD shaft making 1450Nm of torque, and I add a 7in pulley/sheave to the motor shaft. Would the torque change? Half of me says yes other half of my brain thinks no. Torque = Force * distance 1450Nm/(0.075m/2)=19333.3N at the edge of the motor shaft. Now at the 7in Pulley that is bolted to the shaft 19333.3N*0.0889m= 1718.73Nm (0.0889m> 3.5in>half of 7in pulley) Is this right? or would you still have the 1450Nm of torque at the 7in pulley? 


#2
Oct1811, 05:56 PM

P: 9

I think you would still have 1450Nm at the 7inch pulley. i am saying this cause the axis of the shaft and the pulley are the same.
I am a little confused though abt the calculation that u did with torque = force*distance. in a torque a couple acts and in a moment a force acts. the calculation would hav been correct if u were calculating moment, but as u are calculating the torque i think u should multiply the diameter instead of the radius. its like turning a nut with a wrench; a couple is formed between the normal and the force applied and u multiply it with the distance between the two that is the length of the wrench. here the couple is formed inside the motor in the coils with the dia as the distance between them. now when you put a pulley on the shaft, no extra force acts and the same torque acts at the axis of the pulley. its like in an engine, the torque of the engine gets transferred to the tire. the doubt tht i have is if the torque would be effected by the weight of the pulley and the belt you would put on it later cause your talking about a practical problem. (in theory they usually asume the masses as negligible.) 


#3
Oct1911, 10:47 AM

P: 6

I had completely forgotten about Torque copules.
From Wikipedia: A Couple is a system of forces with a resultant (a.k.a. net, or sum) moment but no resultant force. Another term for a couple is a pure moment. Its effect is to create rotation without translation, or more generally without any acceleration of the centre of mass. So, this is exactaly what and electric motor is a pure moment. I guess what keeps blowing my mind is if I look at this assembly like the attached picture. If I apply force F1, large enough to sall the motor, then the force required to stall it will be less at F2. Therefore, the force at F2 would be what is aviable to the belt and next pulley. Thought comment suggestions on this? Oh I noticed a error in my previous calculation 38666.67N*0.0889m=3437.5Nm 


#4
Oct1911, 04:04 PM

P: 190

Pulley Torque calculations
I may be oversimplifying here but...
No, the torque doesn't change. You have 1450 Nm of turning force  whether it is applied over a small or large distance, it remains 1450 Nm. Think of a bolt that requires 100 Nm of tightening torque. You have a socket set with 2 bars, one 1.0 m long and one 0.5 m long. To apply the same torque to the bolt, the 1.0 m bar requires 100 N applied to the end. The 0.5 m bar requires 200 N applied. In your example, the torque applied by your motor to a small shaft or a large pulley will be the same. 1450 Nm. The resultant force will not  it will vary in proportion to the radius. A shaft of (e.g.) 0.1 m will produce a tangent force of 14,500 N (1,450 N / 0.1 m) and a pulley of (e.g) 1.0 m will produce a tangent force of 1,450 N (1,450 N / 1.0 m). It may help to imagine your shaft and pulley as 2 different levers welded to the motor spindle. One is 37.5 mm long (the shaft) and one is 88.9 mm long (the pulley). Clearly the shorter lever will need more force to stall the motor. From http://en.wikipedia.org/wiki/Newton_metre 


#5
Oct1911, 10:12 PM

Mentor
P: 22,313

You're not thinking you can amplify power with a pulley, are you...?



#6
Oct2011, 08:52 AM

P: 6

Bandit127, you hit on my though exactly with your example Because, I keep thinking about it in that way. As a 3.5in(88.9mm) lever driving the belt, therefore I want to take the resultant force from the 7in motor pulley and multiply it by the radius of the 34in pulley to determine the torque applied at the larger pulley. (7in and 34in make up stage 1 in my belt drive) 1450Nm/0.0889m=16310.5N leaving the 7in motor pulley. Torque at the 34in pulley is 16310.5N*0.4318=7042Nm Instead of doing 1450NM (34/7)= 7042Nm Which makes the assumption that the resultant force is equal. Ok, Light bulb moment, I see the error in my thinking. In my initial question I was trying to calculate a different torque at the motor pulley, which cannot happen, but I can have a different resultant force at the larger radius. So continuing on through the drive train, pulley #2 (34in one from above) and pulley #3 (10in OD) are rigidly fixed on the same axle/jackshaft. Therefore torque in = torque out, power in=power out and rpm in =rpm out. We know the torque from above at the 34in pulley is 7042Nm. The resultant force at Pulley #3 is 7042Nm/0.127m=55448.8N Moving out to Pulley #4 with a 33in od. The torque is 55448.8N*0.4191=23238.6Nm (7042Nm*(33/10)=23238.6Nm) Thanks guys for all your help. 


#7
Oct2311, 02:08 PM

PF Gold
P: 244

However, as torque is measured at 1m radius, it would not change in your example. Vidar 


#8
Oct2411, 11:02 AM

P: 6

Vidar,
Could you go into a little more detail? Also, torque is in Nm not N/m so I'm confused by your example. You used the motor torque of 1450Nm divided by the moment arms (0.0375m, 0.0889m) to calculate the resulant force for each So basically (1450Nm/0.0375m)/0.0375m=1031111N/m Where is the 1m radius coming from? 


#9
Oct2411, 02:15 PM

PF Gold
P: 244

If your motor have a shaft of only 75mm, the radius is only 37.5mm, or 0.0375m. The force at the shafts circumference would therfor be 1450Nm/0.0375m which is 38666.666N. The greater the circumference is, the less force will apply to it. Hope my corrections helped you out. Vidar 


#10
Apr2913, 06:23 PM

P: 1

lets simplify this
HP*5252/RPM=TORQUE FTLBS MOTOR HP 2 MOTOR RPM 1725 2*5252/1725=6.08 FTLBS TORQUE ADD A DRIVING PULLEY 3 INCH AND A DRIVEN PULLEY 5 INCH TO THIS AND THROUGH REDUCTION WE WILL INCREASE TORQUE. 1725*3/5=1035 RPM 2HP*5252/1035RPM=10.14 OR 10 FTLBS TORQUE LESS INEFF. OF PULLEYS MOTOR ETC. 


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