torque question

Thread Starter

tracecom

Joined Apr 16, 2010
3,944
Suppose a motor stalls at 20kg.cm and its shaft is .25" in diameter. What is the maximum load that can be exerted on a cord wound directly on the shaft? (Be nice; remember I am a marketing guy.)
 

studiot

Joined Nov 9, 2007
4,998
Well I assume that the cord is wound round securely at one and and the other hangs down, supporting a pan.

Then weights are added until it stalls and the question is what weight stalls it?

So the shaft has diameter 0.25 inches = 6.35mm = .00635m
So the radius is half this or .003175m.

The cord hangs down with tension T equal to the weight in the pan.

This develops a torque of T x radius which we are told stalls at 20 kgcm = 0.2 kgm

So max T x R = 0.2

T = 0.2/0.003175 = 63 kg force, or 63 x g Newtons force = 618N.

So your motor shaft will wind up a load of just under 63kg mass.
 

Thread Starter

tracecom

Joined Apr 16, 2010
3,944
Well I assume that the cord is wound round securely at one and and the other hangs down, supporting a pan.

Then weights are added until it stalls and the question is what weight stalls it?

So the shaft has diameter 0.25 inches = 6.35mm = .00635m
So the radius is half this or .003175m.

The cord hangs down with tension T equal to the weight in the pan.

This develops a torque of T x radius which we are told stalls at 20 kgcm = 0.2 kgm

So max T x R = 0.2

T = 0.2/0.003175 = 63 kg force, or 63 x g Newtons force = 618N.

So your motor shaft will wind up a load of just under 63kg mass.
Thanks for the answer and for showing the math to get there.
 

KL7AJ

Joined Nov 4, 2008
2,229
Suppose a motor stalls atThe 20kg.cm and its shaft is .25" in diameter. What is the maximum load that can be exerted on a cord wound directly on the shaft? (Be nice; remember I am a marketing guy.)
The tangential force is the torque divided by the radius. .125"=.3175cm. 20/.3175=62.9 Kg.

Eric
 

studiot

Joined Nov 9, 2007
4,998
The only real thing is that kilograms force are a non standard unit.

In the ISO metric system the kilogram is a unit of mass and the unit of force that corresponds is the newton.

So working is kgcm is definitely likely to lead to trouble.

since you had a 1/4 inch shaft and are in America, why not keep it imperial and use poundsinches poundsfeet?
 

Thread Starter

tracecom

Joined Apr 16, 2010
3,944
The only real thing is that kilograms force are a non standard unit.

In the ISO metric system the kilogram is a unit of mass and the unit of force that corresponds is the newton.

So working is kgcm is definitely likely to lead to trouble.

since you had a 1/4 inch shaft and are in America, why not keep it imperial and use poundsinches poundsfeet?
I converted 63 kg to 139 pounds, which tells me what I need to know, which is that a 100 pound test cord would be suitable.
 

studiot

Joined Nov 9, 2007
4,998
Glad it is sorted for you.

I assume you know that if you wrapped the cord several times around the shaft the anchoring force needed at the fixed end would be a lot less than the lifting force available at the free end?
 
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