Net torque on a cylinder by multiple applied forces

In summary: The line of action of F4 seems to pass exactly over the rotation center, being the length of its lever
  • #1
Mohmmad Maaitah
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Homework Statement
Find the Net torque on a cylinder.
Relevant Equations
##\tau## = ## \mathbf {Fdsin θ}##

Why did he give torque number 4 zero?

It's not touching the axis of rotation and the angle 90 degrees between them.
I get this:

##\tau## = 1 - 0.8 - 0.4 + 0.4 = 0.2 (C.C.W)

1685629938939.png

1685629956967.png
 
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  • #2
@Mohmmad Maaitah

Do you not believe its perhaps appearing a little impolite that you did not consider taking a moment to try and implement LaTeX Guide?
 
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  • #3
erobz said:
@Mohmmad Maaitah

Do you not believe its perhaps appearing a little impolite that you did not consider taking a moment to try and implement LaTeX Guide?
Yes I know I'm sorry, I would love to use it and I see it's good but I don't remember if I have to based on the forum rules.
I am running out of time currently and I try to save most of it so I didn't check it.
 
  • #4
Mohmmad Maaitah said:
Yes I know I'm sorry, I would love to use it and I see it's good but I don't remember if I have to based on the forum rules.
I am running out of time currently and I try to save most of it so I didn't check it.
Well, your job is to make it as easy for you to receive free homework help here, by making it easy for helpers to interpret and quote your workings. Do think we like wasting time repeatedly asking you this simple courtesy, instead of getting down to the "helping"?
 
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  • #5
erobz said:
Well, your job is to make it as easy for you to receive free homework help here, by making it easy for helpers to interpret and quote your workings. Do think we like wasting time repeatedly asking you this simple courtesy, instead of getting down to the "helping"?
You're right, I will do that next time for sure.
 
  • #6
Mohmmad Maaitah said:
You're right, I will do that next time for sure.
Ever hear the expression "There is no time like the present"? You could have already been finished and on your merry way by now. Instead, you are going to wait for someone that isn't trying to prove a point.
 
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  • #7
Mohmmad Maaitah said:
I see it's good but I don't remember if I have to based on the forum rules.
Yes it is. We sometime cut some slack for folks new to PF to give them time to learn it, but especially for simple equations it is very easy to use:
PF Rules said:
Please clearly state what you wish to discuss. In general, one should attempt to flesh out questions and arguments adequately enough that readers will have a good understanding of the issue. Core post content should be typed out and not solely contained in an image, PDF, or other attachment. (Exceptions at Mentor's discretion.) Use images for supporting figures. Use LaTeX to type mathematical expressions and equations. (See: Learn LaTeX for Math Equations)

There is a helpful "LaTeX Guide" link below the Edit window to get you started. Note that you put double-$ delimiters at the start and end of each stand-alone line of LaTeX, and double-# delimiters at the start and end of in-line LaTeX that does not need to be on its own line. Also, if you right-click on a LaTeX equation in a post, you get a pop-up menu to let you view the LaTeX source or view it in other formats.

LaTeX isn't supported in thread titles, so you can use simple text math in titles if you want.

Note also that PF uses a feature called "lazy LaTeX rendering" that speeds up page loads. When you first post your LaTeX in a thread, you will not see it rendered that first time. Just refresh your browser page to force it to be rendered, and then it should render whenever you come back to that page/thread in the future.
 
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  • #8
To respond to your question. Draw the position vector ##r_4## from the center of the circle to the point of application of force ##F_4##. What is ##F_4r_4\sin\!\theta?##
 
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  • #9
kuruman said:
To respond to your question. Draw the position vector ##r_4## from the center of the circle to the point of application of force ##F_4##. What is ##F_4r_4\sin\!\theta?##
Oh that's clear now
But for the first torque it doesn't look like 90 degree to me between the radius and the action force
 
  • #10
Mohmmad Maaitah said:
Oh that's clear now
But for the first torque it doesn't look like 90 degree to me between the radius and the action force
The drawing is misleading.

If wrapped strings or ropes are represented, forces 1, 2 and 3 can’t be other than tangent to the surfaces of the cylinders; therefore, perpendicular to its lever respect to the axis of rotation, which length must be its corresponding radius.

The line of action of F4 seems to pass exactly over the rotation center, being the length of its lever equal to zero.
 
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  • #11
Mohmmad Maaitah said:
for the first torque it doesn't look like 90 degree to me between the radius and the action force
It's the way the horizontal dashed line has been drawn. In principle , where exactly it is drawn does not matter because it is only there to show the angle to the horizontal (37°), but it gives the impression of having been drawn from the point of contact, which it is not. Imagine it moved up a little.
 
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  • #12
Moment summation.jpg
 
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  • #13
Thanks guys really appreciate all the help I got on this site.
Got 100/100 this semester in physics 101
hope the same for Physics 2!
 
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  • #14

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