Weight required to hang straight down with known torque

In summary, the weight needs to be enough to keep it suspended in the air, but not so much that it breaks the mechanism.
  • #1
MoMan
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How much Weight (W) is required to keep a weight hanging straight down when the torque is known on a rotating shaft? Please see attached image. Assume a lever weight of zero to keep it simple.
Weight calculator.jpg
 
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  • #2
MoMan said:
it simple

Yeah, ##\infty##
 
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  • #3
MoMan said:
How much Weight (W) is required to keep a weight hanging straight down when the torque is known on a rotating shaft? Please see attached image. Assume a shaft weight of zero to keep it simple.View attachment 303451
Suppose that the shaft is deflected so that it does not hang straight down. Suppose that it is deflected by ##x## meters rightward, for instance. Do you know how to calculate how much restoring torque results from gravity pulling on the deflected weight?

There is a simple answer to the question you ask. But the Physics Forums way is to guide you into discovering that answer for yourself.
 
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What's the application ?
 
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The way to answer this question with infinities: The torque of weight around the central point of the shaft is ##T_W=W\cdot 1m\cdot \sin 0##. For any finite W this equals to zero and hence the total torque will be equal to the torque from shaft and it will rotate the weight.
However if ##W=\infty## then $$T_W=\infty\cdot 0=\text {maybe something finite and equal to -}T_{shaft}$$ and hence for infinite weight the system might not rotate.
 
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hmmm27 said:
What's the application ?
It's either a puzzle (and not a very good one, IMO), or he's just trolling.
 
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At what point do macro scale accelerations become imperceptible? Can we not effectively consider the rotation of ##0.1 ^{\circ}## over 10 years as effectively no acceleration. We do this kind of thing all the time in our modeling.
 
  • #8
erobz said:
At what point do macro scale accelerations become imperceptible? Can we not effectively consider the rotation of ##0.1 ^{\circ}## over 10 years as effectively no acceleration. We do this kind of thing all the time in our modeling.
Well after the point at which suspended weights break the mechanism.
 
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$$m=\frac\tau{\mu_s{rg}}$$
 
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Related to Weight required to hang straight down with known torque

1. What is the formula for calculating the weight required to hang straight down with known torque?

The formula for calculating the weight required to hang straight down with known torque is W = T / r, where W is the weight in newtons, T is the torque in newton-meters, and r is the distance from the pivot point to the point where the weight is hanging.

2. How do I determine the torque required for a specific weight to hang straight down?

To determine the torque required for a specific weight to hang straight down, you can use the formula T = W * r, where T is the torque in newton-meters, W is the weight in newtons, and r is the distance from the pivot point to the point where the weight is hanging.

3. What is the unit of measurement for torque?

The unit of measurement for torque is newton-meters (N*m).

4. How does the distance from the pivot point affect the weight required to hang straight down with known torque?

The weight required to hang straight down with known torque is directly proportional to the distance from the pivot point. This means that the weight will increase as the distance from the pivot point increases.

5. Can I use this formula to calculate the weight required for any object to hang straight down?

Yes, this formula can be used to calculate the weight required for any object to hang straight down as long as the torque and distance from the pivot point are known.

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