Torque/Moment of a bent wrench question

[NewProgramma]

Hi everyone,
The problem below kinda confuses me...i've spend half an hour staring at this problem lol.

Homework Statement :

Q: A bent wrench is used to apply a moment of 221 Nm to the hexagonal nut. If the applied force F is 507 N and the angle θ is 21°, find the distance d. Answer in metres.

http://imgur.com/pLjSJkQ

http://[noparse]http://imgur.com/pLjSJkQ[/noparse]

https://courses.kpu.ca/pluginfile.php/287327/question/questiontext/205374/1/856670/q1.jpg The attempt at a solution:
Here is what I tried to do:

Assuming: clockwise moment = negative.

First:
break F into components: 507cos21 = Fx and 507sin21 = Fy.

Second:
Since Fy is along the line of 'd' it has zero moment at the "bent pivot" (not at the nut).
But Fy will exert a moment at the hexagonal nut: so break Fy into components: 507sin21sin28.

Third:
find net moment: -(507cos21)*d + (507sin21sin28)*0.3 = -221
So, d = 0.52m

But the correct answer given to us is: 0.26m

I think I might have mad an error with the moment at "the bent pivot". Since "-(507cos21)*d" equals to the moment at the "bent" and not at the Origin(hex nut)...should I shift the moment at the bent to the origin(nut)?
Or I can think of it as part of the net moment?

I'm still a bit confused about the Net moment at the Origin(nut), does the Net moment mean the sum of all the moments acted on the body, regardless of where it acts? Or does all the moments have to be on the Origin?

Thank you all for the generous help.

 

[SteamKing]

Hi everyone,
The problem below kinda confuses me...i've spend half an hour staring at this problem lol.

Homework Statement :

Q: A bent wrench is used to apply a moment of 221 Nm to the hexagonal nut. If the applied force F is 507 N and the angle θ is 21°, find the distance d. Answer in metres.

http://imgur.com/pLjSJkQ

http://[noparse]http://imgur.com/pLjSJkQ[/noparse]

https://courses.kpu.ca/pluginfile.php/287327/question/questiontext/205374/1/856670/q1.jpg The attempt at a solution:
Here is what I tried to do:

Assuming: clockwise moment = negative.

First:
break F into components: 507cos21 = Fx and 507sin21 = Fy.

Second:
Since Fy is along the line of 'd' it has zero moment at the "bent pivot" (not at the nut).
But Fy will exert a moment at the hexagonal nut: so break Fy into components: 507sin21sin28.

Third:
find net moment: -(507cos21)*d + (507sin21sin28)*0.3 = -221
So, d = 0.52m

But the correct answer given to us is: 0.26m

I think I might have mad an error with the moment at "the bent pivot". Since "-(507cos21)*d" equals to the moment at the "bent" and not at the Origin(hex nut)...should I shift the moment at the bent to the origin(nut)?
Or I can think of it as part of the net moment?

I'm still a bit confused about the Net moment at the Origin(nut), does the Net moment mean the sum of all the moments acted on the body, regardless of where it acts? Or does all the moments have to be on the Origin?

Thank you all for the generous help.

This problem is quite simple if you know that M = r × F , where M, r, and F are all vectors and you are taking the cross product of r and F.

If you know only the basic definition of the moment of a force, it's still easy.

Take for example this curly wrench:

http://web.mit.edu/4.441/1_lectures/1_lecture5/wrench1.gif
​

The moment arm is 12" between the red arrow, which indicates the applied force, and the center of the nut or bolt being tightened. The shape of the wrench in between these two points is immaterial to how much torque is generated.

For your particular problem, you need to find the distance between the center of the bolt and the tail of the force vector. Also, you need to figure out the component of the turning force which acts at right angles to this lever arm. Then, you can work back and find the distance d.

 

[haruspex]

The component F sin(21)sin(28) acts through what point? How far is that from the nut?
The component F cos(21) acts in what direction and is how far from the nut?
But an easier way (in my view) is to extend the 30cm line out and drop a perpendicular to it from the tip of the wrech where F is applied. What is the angle between that perpendicular and the line of action of F?

 

[NewProgramma]

Oh, I've tried to attach an image of the wrench, but it doesn't seem to display properly here, so I added the link http://imgur.com/pLjSJkQ
of the diagram.
sry about that.

 

[haruspex]

Oh, I've tried to attach an image of the wrench, but it doesn't seem to display properly here, so I added the link http://imgur.com/pLjSJkQ
of the diagram.
sry about that.

Yes, I followed the link and saw the image. My comments stand.