[mnathan]

Hi,
I would like to calculate the torque of a system,
I have described the system below,

Consider the conveyor belt, the motor rotates the conveyor belt (25 Meters), I want to do a reverse engineering,

Let say the I don't use the motor to run the conveyor, instead I have a handle (1 meter) attached to the belt at one point, if I use my hand force to push the handle which in turn will turn the motor. I want to calculate the torque applied at the motor input when I input some 5N force at the edge of the handle, and let say I have increased the handle length to 2 meters, I want to apply the same force 5N on the edge of the 2 meter handle, does the torque output will be doubled on the motor input .as compared to the 1 meter handle.

Looking forward to hear from you,
Thanks in Advance,
Manjunathan.S

[ebaines]

Torque is calculated by taking the force and multiplying by the length of the lever arm from the center of rotation. Mathematically it's:



Note that torque is a vector and it's the cross product of the distance and force - hence to maximize torque you want the force to be exactly perpendicular to the handle. Intuitively that should seem pretty obvious.

So yes - if you double the length of the lever arm you double the amount of torque (assuming is constant).

[mnathan]

Hi,
I would like to calculate the torque of a system,
I have described the system below,

consider the conveyor belt, the motor rotates the conveyor belt (25 Meters), I want to do a reverse engineering,

Let say the i don't use the motor to run the conveyor, instead i have a handle (1 meter) attached to the belt at one point, if i use my hand force to push the handle which in turn will turn the motor. i want to calculate the torque applied at the motor input when i input some 5N force at the edge of the handle, and let say i have increased the handle length to 2 meters, i want to apply the same force 5N on the edge of the 2 meter handle, does the torque output will be doubled on the motor input .as compared to the 1 meter handle.?

Looking forward to hear from you,
Thanks in Advance,
Manjunathan.S

Hi ebaines,
Thanks for your answer. But the conveyor belt is not a circular shape. Kindly see the right most image on this site Conveyor belt - Wikipedia, the free encyclopedia, instead of rotating the shaft, I need to put as handle on the belt,it is a fixed handle to one point of the belt, and use a force to push the handle which inturn rotates the shaft,
My question is,
If I increase the length of the handle and use same force as the previous handle, as you said then the torque on the shaft will also increase. Am I right?

Thanks once again,
Manjunathan.S

[ebaines]

You need to evaluate the cross product of the two vectors: one being the distance from the center of the motor shaft to the end of your handle, and the other being the force you apply to the handle. If we call the length of the handle as "L" and the radius of the conveyor belt roller R, to double the torque you need to double the value of (R+L). You can see that merely doubling L won't do it.

[mnathan]

Thanks Once Again ebaines,
I have a question, If the handle is in the middle of the belt, does the belt length to the shaft will affect the torque output, what is the difference of torque will I get when let say the handle is close to the shaft, and handle is very far away from the shaft, (consider the left part shaft is used for torque measurement).

Thanks Once Again,
Manjunathan.S

[ebaines]

Makes no difference - since the torque is the cross-product of the two vectors I mentioned before. In the figure below, if forces F1 and F2 are of the same magnitiude and direction, they both apply the same torque to the left roller. Note that they both present the same moment arm of (R+L) to the roller.

[mnathan]

Thanks Once again,
1. If I have two handles and give force F1 and F2 to handle 1 and 2 respectively, will the torque generated will be like this T =(F1 + F2) (R+L) ?

Thanks in Advance,
Have a Great Day,
Regards,
Manjunathan.S

[ebaines]

If i have two handles and give force F1 and F2 to handle 1 and 2 respectively, will the torque generated will be like this T =(F1 + F2) (R+L) ?

Yes.

[mnathan]

Thanks ebanis,

Consider one of the wheel(size =R,Torque=T Nm) is connected to the shaft (size R/2 in diameter), other end of the shaft is connected to a input gear (size=R*6), this again connected to another output gear (size R), from the calculation,at the output gear I get 6 times the wheel rotation, but what will be for the torque conversion, is it at the output gear will I get T/6 Nm or something else.

Consider R is small (let say R=10 cm) compared to the handle length (3 meters),

I have given Force=20 N on the handle, since torque at wheel will be T=20 (3+0.1)=60.2 Nm

What will be the torque at the output gear ?


Thanks Once Again,
Have a Great Day,
Regards,
Manjunathan.S

[ebaines]

The output gear will turn at 6 times the rotational velocity of the input wheel, and consequently will provide 1/6 the torque.

[mnathan]

Thanks Once again Ebanies,
I have another doubt,

1. let say I am holding above the ground a string of 2 meter with 1 Kilogram attached to it, if I use force 1N to pull. So I am able to pull the srting in 1 Meter in 1 sec.

2. Let say I have above said conveyaor belt with attached handle to it (let say handle length is 1 meter) I putting force of 1N on this handle, As per your definition the torque generated on the shaft would be 1N(1meter+R), Consider R is small enough to ignore, the torque generated would be 1 * 1=1Nm. Let say I have a string attached to the shaft, and string (10 meters length) is holding 1Kilogram. When I apply 1N on the handle, will I am able to the pull the string more than 1 meter above the ground, or is it possible to add more kilogram (say 1.2 KG) and pull it to 1 meter in 1 sec. Basically what is the performance difference between the 1 and 2 design? In performance wise will 2 is superior than the 1 ?

Thanks in advance,
Regards,
Manjunathan.S

[ebaines]

Sorry - I don't follow you. Somehow you have a string with a weight that is attached to the handle, and that weight is somehow applying a force to the string? How? And what do you mean by "pull the string more than 1meter above the ground?"

In your item 1 above - if you pull on a 1 Kg weight with 1 N force, it will not move at a constant 1 m/s, but will accelerate at 1 m/s^2, if you ignore friction.

[mnathan]

Hi Ebanis,
My question is what will be the performance,
Consider I give just 1N force as a pulse (only once),

A.
As described in step 1, If I pull the string with the force (a pulse of 1N), I should be able to pull 1 Meter in one second, Let say if the string is holding 2 Kilogram then I would say I can pull ony 1/2 a meter in one second am I right ?


B.
Now let say we have a system as described by the conveyor belt with a handle attached to it, the shaft which is attached to the conveyor belt is holding a string with 1 kilo gram.
Now let say the handle is 1 meter, I apply a force (single pulse) of 1 N on the handle, Will the shaft be able to pull the string, now let say I increase the Kilogram by 2KG, what will happen in that case ?

C.
And let say I have a handle of 2 Meters, now I apply the same force of 1N, does in this case the shaft is able to pull the string as same distance but the twice the Kilogram as section B ?

What is the performance difference between the A and B system?


Thanks in Advance,

Regards,
Manjunathan.S

[ebaines]

Hi Ebanis,
My question is what will be the performance,
Consider I give just 1N force as a pulse (only once),

A.
As described in step 1, If i pull the string with the force (a pulse of 1N), i should be able to pull 1 Meter in one second, Let say if the string is holding 2 Kilogram then i would say i can pull ony 1/2 a meter in one second am i right ?

Yes - from physics of impulses:



So if you give a suffient impulse (F ) to change the velocity of a 1 Kg mass m by 1 m/s, that same impulse will accelerate a 2 Kg mass by half as much.

B.
Now let say we have a system as described by the conveyor belt with a handle attached to it, the shaft which is attached to the conveyor belt is holding a string with 1 kilo gram.
Now let say the handle is 1 meter, i apply a force (single pulse) of 1 N on the handle, Will the shaft be able to pull the string, now let say i increase the Kilogram by 2KG, what will happen in that case ?

It sounds like you want to bump the conveyor with an impulse of sufficient amplitue to change the velocity of a 1Kg mass by 1m/s, and then are wondering what would be the effect of that same magntude of bump if the conveyor has a 2 Kg mass on it? Again, from it seems that the mass would move at 0.5 m/s. You can do the same calculation based on torque as well:

C.
and let say i have a handle of 2 Meters, now i apply the same force of 1N, does in this case the shaft is able to pull the string as same distance but the twice the Kilogram as section B ?

What is the performance difference between the A and B system?

Doubling the moment arm, and givng the same "bump" force, doubles the torque bump, and hence the rotation of the system () doubles.

[mnathan]

Thanks Again,
Is the impulse force given in 1 and 2 is different to make 1 KG move 1 meter in 1 sec ?
Which one requires more force ?

And another thing,
Let say I have handle 1 meter, I give some 'X' N impulse force, and let say object moved Y Meter/Sec, now I reduced the handle by 0.5 meter, I apply the same impulse force 'X' N, now the object moves Y/2 Meters/Sec, now I reduce again the handle to 0.1 then apply same impulse force 'X' N, now the object moves Y/10 meters per sec am I right ?


Not let say I directly hit the object with same impulse force 'X' N, how much distance the object will move,

Consider in all cases the friction is same.

Thanks in Advance,
Regards,
Manjunathan.S

[Unknown008]

Let say i have handle 1 meter, i give some 'X' N impulse force, and let say object moved Y Meter/Sec, now i reduced the handle by 0.5 meter, i apply the same impulse force 'X' N, now the object moves Y/2 Meters/Sec, now i reduce again the handle to 0.1 then apply same impulse force 'X' N, now the object moves Y/10 meters per sec am i right ?

Yup, these two are directly proportional, so, you're right.

But let's wait for ebaines answer for the next question. He knows better. :)

[ebaines]

Thanks Again,
Not let say I directly hit the object with same impulse force 'X' N, how much distance the object will move,?

It should be clear to you by now that rotating objects accelerate their rate of rotation based on the torque applied. Torque is the force you apply multiplied by the length of the moment arm of that force, which means the distance from the object's center of rotation to the point where the force is applied. More accurately - it's actually the cross product of the force vector and the vector from the center of rotation to the applied force. So yes, if you make the moment arm 0 length, you are applying 0 torque regardles of how hard you push. In your conveyor belt contraption the moment arm is measured from the center of the rollers that the belt winds around.

[mnathan]

Thanks Ebanies,
But I have a question, if I hit the object (IN THIS CASE THERE IS NO CONVEYOR BELT OR SHAFT OR HANDLE, i.e. FORCE DIRECTLY HITTING THE OBJECT) directly with force it should move from its original position right?
How much distance will it move compared to the system having conveyor belt,shaft,handle(1meter) ?

When apply the same force, Will the object move better distance in directly hitting or object will move better distance when used conveyor belt,handle,shaft ? Which one is better ?

Thanks in advance
Regards,
Manjunathan.S

[ebaines]

It's impossible to compare these two diffferent systems unless you provide some additional data:

1. For the version where the object is just going to slide across the floor - what is the coefficient of friction between the object and the floor?
2. For the conveyor version, what are the masses and moment or inertia of the conveyor belt and rollers? Can one assume frictionless rollers?

If one can assume that in version 1 friction is 0 (i.e. the object is sliding on ice), and in version 2 that the conveyor belt and roller have negligible mass and no friction in the roller mechanism, then the 2 cases turn out to be the same. In both cases it comes down to

[mnathan]

Thanks Ebanies,
So if I use 2 meter handle then I can say that conveyor belt system is better than the directly hitting the object, since here torque is doubled. Am I right ?

Thank in Advance
Regards,
Manjunathan.s