I am presenting a solution to a physics problem to some high school students in my robotics club. I haven’t taught physics for a while so I wanted to check with straight dope experts to make sure I did the math right.
Here is the problem.
A mechanical arm supports 4 kg of water which is located a distance (d) from the fulcrum.
According to the motor specs. the stall torque is 198.26 oz-in. I converted this to 221 Nm. Is this correct?
Using torque = Fd found that without any mechanical advantage, the max distance the arm can be to lift the water is 35 cm. Is this correct?
Assume a gear ratio of 10:1, the arm can now be 350 cm long. Is this correct?
Your figure of 221 Nm can’t be correct, and I’m not sure how you got it, since a newton is more than an ounce, and a meter is much more than an inch. I get 1.4 Nm, with enough zeroes after the 1.4 that I suspect that that’s the motor’s actual rating, and that 198.26 is an overly-precise conversion.
You apparently didn’t use your figure of 221 Nm in the next step, but instead somehow re-did the conversions, because 35 cm is the length of arm to get 1.4 Nm from a 4 newton force. But that’s still not right, because you have 4 kg of water, not 4 N. The weight of 4 kg of water will be nearly 40 N (more precisely, 39.2 N), so your actual answer will be closer to 3.5 cm.
And that’s also assuming that the arm is perfectly horizontal, because torque isn’t just force times distance, it’s force times distance times the sine of the angle between them. Without more detail, I don’t know if this is what you want or not.
And crap, you’re right about the 3.5 cm. I feel really dumb. I converted 0.035 m to 35 cm.
Now, I’m probably wrong on the mechanical advantage part at the end.
I think I would want a gear ratio of 10:1 to get to increase the length of the arm 10X.
You’re still being imprecise. Which I suspect is due to hurry. One can’t successfully teach anything in a hurry. The inevitable mistakes really buffalo the students.
Is the 10:1 gear ratio you’re thinking of from motor:lever arm or from lever arm:motor?
Bottom line: If you want 10x the lever arm you’ll need the motor shaft to turn 10x the revs of the lever arm shaft.
You’ll also need some margin. If the motor stalls at 1.4 Nm you’ll want to plan to extract no more than ballpark 90% of that, so 1.25 Nm. Otherwise operation becomes slow and erratic.
You’ll also need to allow for friction losses in any gear train. WAG another 10% for that, plus a small additional factor for each additional gear in the train above two. And another factor for whether you’re using high quality, or cheap-ass, or even homemade gears.
I would have guessed that, for a robotics club, he’s using Lego gears, which are pretty high-quality for plastic, except that 10:1 isn’t an easy gear ratio to get from those (the largest and smallest standard gears are 40 teeth and 8).
Not at all. We build big robots, with elaborate drive trains etc. and compete in FIRST competitions. My son just joined. There is a lot to learn.
Thanks for everyone’s help.
You could take it that the arm will move through horizontal , and horizontal arm is where the motor must produce most torque, as moving the water horizontally takes (toward) 0 force… ( a speed of a nanometer a millenia is implied to be sufficient to prevent motor stalling… Not in practice but the OP didn’t specificy a minimum speed… )
