Finding the Torque of a shaft that is spun by a lever

In summary: B compared to shaft A.Yes, as the shorter lever will have less effort force applied, the torque at the shaft will be twice as high.
  • #1
Muhsin
5
0
Homework Statement
Hi there.

I am new to this forum, but I joined just to ask some input from any one who may be kind to help.

I have this attached file where you can see all the values and equations I have used but there is one bit where I need help. finding Torque on Shaft B, I appreciate if someone could advice how to convert the load force into Torque on the shaft.

if anything I have done it is not correct please feel free to advice and share your input.

Thanks
Relevant Equations
Torque = Force * Radius
Homework Statement: Hi there.

I am new to this forum, but I joined just to ask some input from anyone who may be kind to help.

I have this attached file where you can see all the values and equations I have used but there is one bit where I need help. finding Torque on Shaft B, I appreciate if someone could advice how to convert the load force into Torque on the shaft.

if anything I have done it is not correct please feel free to advice and share your input.

Thanks
Homework Equations: Torque = Force * Radius

Torque = Force * Radius

Effort Force * Lever length = Load Force * arm Length
 

Attachments

  • Mechanical lever.jpg
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  • #2
It is hard to check your work without being able to distinguish between the original problem and your own work. Please post the original problem as given to you.
One hint: consider the work done in one revolution. How does that relate to the torque at each end?
 
Last edited:
  • #3
Muhsin said:
Homework Statement: Hi there.

I am new to this forum, but I joined just to ask some input from anyone who may be kind to help.

I have this attached file where you can see all the values and equations I have used but there is one bit where I need help. finding Torque on Shaft B, I appreciate if someone could advice how to convert the load force into Torque on the shaft.

if anything I have done it is not correct please feel free to advice and share your input.

Thanks
Homework Equations: Torque = Force * Radius

Homework Statement: Hi there.

I am new to this forum, but I joined just to ask some input from anyone who may be kind to help.

I have this attached file where you can see all the values and equations I have used but there is one bit where I need help. finding Torque on Shaft B, I appreciate if someone could advice how to convert the load force into Torque on the shaft.

if anything I have done it is not correct please feel free to advice and share your input.

Thanks
Homework Equations: Torque = Force * Radius

Torque = Force * Radius

Effort Force * Lever length = Load Force * arm Length

Thanks for a quick reply.

The task was for me to prove if the lever can amplify the torque on shaft B even though Shaft A and Shaft B have the same rpm.

So, since lever amplify linear force in mechanical advantage. Will it produce the same advantages if it is used in the design used in the attached? If it does based on all given parameters what would be the torque in shaft B?

By the way this has been used in a windmill design.

Thanks
 
Last edited:
  • #4
Muhsin said:
The task was for me to prove if the lever can amplify the torque on shaft B even though Shaft A and Shaft B have the same rpm.
In that case, although the hint I gave is the quickest way to solve it, it sounds like they want you to go the long way round.
In the diagram you have a cross-piece length 2x50cm. Each end is attached to a lever arm distance 200cm from a fulcrum and continuing 20cm beyond to connect to another cross-piece.

Try to answer these questions in sequence:
How long is the second cross-piece?
If the applied torque is τ, what force is exerted at the ends of the first cross-piece?
What force does that result in at the second cross-piece?
What torque does that generate on the load?
 
  • #5
I already done the follwoing but was not sure if I have done some mistakes.

First converting Torque to Linear Force
Torque = Force * Radius
F = 10 N.m/0.5 = 20 N

since both levers are connected to the same arm the force will be devided between both.
each lever will have around 10N of effort force

now finding load force at the other end of the lever.

Total Effort force * lever length upto the Fulcrum = Load force * continuing lever length
20 N * 2 m = Fl * 0.2
Fl = 40 / 0.2 = 200 N

The linear force on each continuing lever length is 100N

Now, we convert the linear force on the 10cm length lever on the shaft B to Torque assuming the distance of the end of the continuing lever to the center of Shaft B is 10cm

Torque = 200 * 0.1 = 20N torque of shaft B assuming there is no loss due to friction and other factors

if what I have done above is true. that means the torques are equal. but if I had a shorter contiuing lever length of 10cm instead of 20cm that means the torque will be twice on shaft B compared to shaft A.

is that correct?
 
  • #6
Muhsin said:
that means the torques are equal.
Yes.
Muhsin said:
if I had a shorter contiuing lever length of 10cm instead of 20cm that means the torque will be twice on shaft B compared to shaft A.
No.
If the continuing arm were halved in length then span of the cross-piece attached to it would also halve.
 
  • #7
what if the design changed to the one in the attached i have just sent. won't the torque increase by 50% or 100% ?
or which length i should calculate if i use the second design? the 5cm contiuing lever legnth or the 10cm contiuing lever length or both have to be added then used in the equation?
 

Attachments

  • Mechanical lever 1.jpg
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  • #8
Muhsin said:
what if the design changed to the one in the attached i have just sent. won't the torque increase by 50% or 100% ?
or which length i should calculate if i use the second design? the 5cm contiuing lever legnth or the 10cm contiuing lever length or both have to be added then used in the equation?
I don’t see how changing the shape of the arms alters anything.
Consider my hint in post #2.
 
  • #9
The work will be the same on both end hence the RPM are the same. but in the design we have a lever since most of the time the force requires to move the arm on shaft B is always less than the effort force due to the fulcrum taking some of the load it should increase the torque.

i may be worng but these two design mix between lever and rotational movement causes some confusion for me.
thank you anyway for the support. I guess the best way to find out is to design it and test it.
 
  • #10
Muhsin said:
The work will be the same on both
Yes.
Muhsin said:
the RPM are the same
Yes, but not because the work is the same. The rpm will be the same as a result of the linkage.

What equation relates torque to work done?
 

1. What is torque and how is it related to a shaft and lever?

Torque is a measure of a force's ability to rotate an object around an axis. In the case of a shaft spun by a lever, torque is the force that is applied to the shaft by the lever, causing it to rotate.

2. How do you calculate the torque of a shaft spun by a lever?

The torque of a shaft spun by a lever can be calculated by multiplying the force applied by the lever by the distance between the point of application and the axis of rotation.

3. What factors affect the torque of a shaft spun by a lever?

The torque of a shaft spun by a lever is affected by the force applied by the lever, the distance between the point of application and the axis of rotation, and the angle at which the force is applied.

4. How is the torque of a shaft spun by a lever measured?

The torque of a shaft spun by a lever can be measured using a torque wrench or by using mathematical equations to calculate the torque based on the force and distance applied.

5. How can the torque of a shaft spun by a lever be increased?

The torque of a shaft spun by a lever can be increased by increasing the force applied by the lever, increasing the distance between the point of application and the axis of rotation, or by applying the force at a greater angle. However, it is important to note that there may be limitations based on the strength and design of the shaft and lever.

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