Calculating Tangential Force for Grindstone Rotation

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In summary: \tau_{friction} = \tau_{required}$$you must have more torque applied to the handle than to the axle in order to move the stone.
  • #1
henry3369
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Homework Statement


A 50.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 160 N (Fig. P10.57). The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50 N*m between the axle of the stone and its bearings. How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to in 9.00 s?

Homework Equations


Net torque = Iα

The Attempt at a Solution


α = 1.396 rad/s^2
Torque from axle + Torque from friction + Torque from crank handle = Iα
6.50 + (0.6)(.26)(160) - thandle = (1/2)(50)(.26)2(1.396)
thandle = 29.1 N*m = rF
F = 29.1/.26 = 111.926 N

The answer is 67.6 N though. What am I doing wrong?
 
Last edited:
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  • #2
You want to rethink how the various torques add up to yield the necessary total torque. Which ones add and which ones subtract from the net?
 
  • #3
gneill said:
You want to rethink how the various torques add up to yield the necessary total torque. Which ones add and which ones subtract from the net?
Aren't the signs on the individual torques correct? The two signs of torques from friction will be the opposite from the torque from the handle because friction opposes motion, which is what I have (handle is negative because it rotates it CW, while the torque from friction wants to rotate it CCW)
 
  • #4
The torque from the handle must overcome both the axle friction and the axe friction and leave enough left over to yield the required total.
 
  • #5
I don't understand how this helps with the problem. I know that the net torque has to be greater than zero because the torque from the handle must over the two frictional torques. Isn't that what the equation I used means?
 
  • #6
I can't find net torque (the torque left over to create the acceleration), unless I find torque from the handle first, which is what I'm solving for.
 
  • #7
gneill said:
The torque from the handle must overcome both the axle friction and the axe friction and leave enough left over to yield the required total.
Sorry, I can find net torque from Iα, which I did. And I'm solving for the torque from the handle. My equation is saying that there is more torque going CW because otherwise it would be in equilibrium.
 
  • #8
henry3369 said:
I don't understand how this helps with the problem. I know that the net torque has to be greater than zero because the torque from the handle must over the two frictional torques. Isn't that what the equation I used means?
You wrote:
henry3369 said:
6.50 + (0.6)(.26)(160) - thandle = (1/2)(50)(.26)2(1.396)
which translates to:
$$\tau_{axle} + \tau_{friction} - \tau_{handle} = \tau_{required}$$
Surely the handle torque must exceed the frictional torques, otherwise the grindstone would rotate backwards. Further, the frictional torques should subtract from the handle torque, with what remains giving the net required torque.
 
  • #9
gneill said:
You wrote:

which translates to:
$$\tau_{axle} + \tau_{friction} - \tau_{handle} = \tau_{required}$$
Surely the handle torque must exceed the frictional torques, otherwise the grindstone would rotate backwards. Further, the frictional torques should subtract from the handle torque, with what remains giving the net required torque.
The frictional force causes it to rotate in the counter-clockwise direction though, which means it should be positive right? Why would you subtract it then?
 
  • #10
henry3369 said:
The frictional force causes it to rotate in the counter-clockwise direction though, which means it should be positive right? Why would you subtract it then?
Also, isn't τrequired the same as net torque?
 
  • #11
henry3369 said:
Also, isn't τrequired the same as net torque?
I wrote "required" in order to highlight the fact that it is the amount of torque needed to achieve the desired acceleration. Perhaps I should have left it as "net". Sorry for any confusion that may have caused.
 
  • #12
gneill said:
I wrote "required" in order to highlight the fact that it is the amount of torque needed to achieve the desired acceleration. Perhaps I should have left it as "net". Sorry for any confusion that may have caused.
τnet = Iα
τnet = (1/2)(50)(.26^2)(1.396) = 2.35924 N*m
τnet = τaxle, friction + τaxle, friction - τhandle
τhandle = τaxle, friction + τaxle, friction - τnet
τhandle = 6.50 (given) + μnR (torque due to friction from the axe) - 2.35924 (found above)
τhandle = 6.50 + (0.6)(160)(0.26) - 2.35924
τhandle = 29.10 Nm
τhandle = perpendicular force * distance from the axis = 29.10Nm
FR = 29.10
F = 29.10/0.26 = 111.80 N

What am I doing wrong? I still haven't gotten the answer... My work shows that I am solving for the torque of the handle. I've done this step by step many times and always end up with the same answer.
 
  • #13
You can choose your CW/CCW positive/negative convention either way, but in the end you have one torque that is providing impetus in the desired direction and two that are opposed. Thus:

$$\tau_{handle} - \tau_{axle} - \tau_{axe} = \tau_{net}$$

Note that the net torque has the same sign as the handle torque.

Adjust the signs of all terms by multiplying through by -1 if you wish.
 
  • #14
gneill said:
You can choose your CW/CCW positive/negative convention either way, but in the end you have one torque that is providing impetus in the desired direction and two that are opposed. Thus:

$$\tau_{handle} - \tau_{axle} - \tau_{axe} = \tau_{net}$$

Note that the net torque has the same sign as the handle torque.

Adjust the signs of all terms by multiplying through by -1 if you wish.
This doesn't change my results though. It just changes the sign. I'll still end up with a magnitude of 110.80, which is incorrect.
 
  • #15
I'm trying to figure out why the magnitude of the the force is different, not the direction.
 
  • #16
henry3369 said:
How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to in 9.00 s?
Is there some information left out here?

Does the stone have a CW or CCW angular acceleration α? Did you consider the sign of α?

Also, did you use the correct radius for the crank in your calculation?
 
  • #17
TSny said:
Is there some information left out here?

Does the stone have a CW or CCW angular acceleration α? Did you consider the sign of α?
Oh. For some reason it didn't show up when I copied and pasted it. The final angular velocity is 120 rev/min = 4pi rad/s. I still included the acceleration above though.
 
  • #18
Let's take a look using the magnitudes of the known torques. You have the following known numbers:

Taxle = 6.50 Nm
Taxe = 24.96 Nm
Tnet = 2.359 Nm

and one unknown, Thand

You need Thand to exceed Taxle + Taxe by the amount Tnet. That is,

Tnet = Thand - (Taxle + Taxe)

Thand = Tnet + Taxle + Taxe
 
  • #19
henry3369 said:
Oh. For some reason it didn't show up when I copied and pasted it. The final angular velocity is 120 rev/min = 4pi rad/s. I still included the acceleration above though.
Is the angular acceleration positive or negative?
 
  • #20
TSny said:
Is the angular acceleration positive or negative?
Negative. That would only increase the torque of the handle above 111 though.
 
  • #21
Did you use the correct radius for the torque of the crank handle?
 
  • #22
gneill said:
Let's take a look using the magnitudes of the known torques. You have the following known numbers:

Taxle = 6.50 Nm
Taxe = 24.96 Nm
Tnet = 2.359 Nm

and one unknown, Thand

You need Thand to exceed Taxle + Taxe by the amount Tnet. That is,

Tnet = Thand - (Taxle + Taxe)

Thand = Tnet + Taxle + Taxe
This would result in τhandle = 130.15 N*m, which is still incorrect.
 
  • #23
TSny said:
Did you use the correct radius for the torque of the crank handle?
Well, the crank handle is 0.25 meters away from the axis of rotation, it also gave the length of the handle to be 0.500 m. Does this contribute to the torque at all?
 
  • #24
It would be nice to have a picture. I was assuming the crank handle is .500 m from the axis of rotation. That appears to give the correct answer. But, if it is actually .25 m, then I don't see how to get the answer.
 
  • #25
http://imgur.com/M2jitam

Exact problem. I'm solving for part a. The radius is 0.520/2 = 0.26
 
Last edited by a moderator:
  • #26
henry3369 said:
This would result in τhandle = 130.15 N*m, which is still incorrect.
How does 2.359 + 6.50 + 24.96 equal 130.15?
 
  • #27
gneill said:
How does 2.359 + 6.50 + 24.96 equal 130.15?
Sorry I mean't that equals 33.819 N*m.
Then 33.819/.26 = 130.15 N of force.
 
  • #28
I figured out that using 0.500 m does solve the problem. I was confused when the problem said the handle was 0.500 meters because I thought that he meant that the part of the handle that was perpendicular to the grindstone was 0.500 meters while the length of the handle was 0.26 meters.

Thank you!
 

Related to Calculating Tangential Force for Grindstone Rotation

1. What is torque?

Torque is a measure of the force that causes an object to rotate about an axis. It is calculated as the product of the force applied and the distance from the axis of rotation to the point where the force is applied.

2. How does torque relate to "grindstone and axe"?

In the context of "grindstone and axe", torque refers to the rotational force that is applied to the grindstone in order to sharpen the blade of the axe. The greater the torque, the more effective the grinding process will be.

3. What factors affect the amount of torque produced?

The amount of torque produced depends on the magnitude of the force applied, the distance from the axis of rotation to the point where the force is applied, and the angle at which the force is applied.

4. How can torque be increased?

To increase the torque, you can either increase the force applied or increase the distance from the axis of rotation to the point where the force is applied. Alternatively, changing the angle at which the force is applied can also increase the torque.

5. What is the unit of measurement for torque?

The unit of measurement for torque is the Newton-meter (Nm) in the SI system. In the imperial system, it is measured in foot-pounds (ft-lb) or pound-force feet (lbf-ft).

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