Moment about a drum (Answer Check)

  • Thread starter Thread starter jegues
  • Start date Start date
  • Tags Tags
    Drum Moment
Click For Summary

Homework Help Overview

The discussion revolves around calculating the moment about a point related to a drum system, involving forces and angles. The original poster attempts to verify their calculation of the angle alpha based on the moment equation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods of calculating moments, including breaking forces into components and considering the geometry of the system. Some question the accuracy of the original poster's angle calculation and explore alternative approaches to verify the results.

Discussion Status

There is ongoing dialogue about the correctness of the calculations, with some participants suggesting different methods to check the results. The original poster is seeking confirmation of their findings, while others are exploring discrepancies in the calculations.

Contextual Notes

Participants note the importance of the forces' alignment with the center of the drum and the implications of relocating forces in the analysis. There is mention of differing results for the angle alpha, indicating potential misunderstandings or errors in calculations.

jegues
Messages
1,085
Reaction score
3

Homework Statement


See Figure.


Homework Equations


[tex]M = Fd[/tex]


The Attempt at a Solution



I split all the forces into rectangular components and calculated their perpendicular distance from the point E.

[tex]M_{R}= 50 = (15.5 * 0.72) + (58 * 0.56) - (63.1 * 0.61) + (90.1 * 1.18) - (130sin \alpha * (0.75 + 0.75cos \alpha ) ) + (130cos \alpha * 0.75sin \alpha)[/tex]

So,

[tex]\alpha = 39.1^{o}[/tex]

Can someone verify my answer by chance?

Thanks again.
 

Attachments

  • DrumQuestion.JPG
    DrumQuestion.JPG
    30.9 KB · Views: 409
Physics news on Phys.org
Bump still looking for an answer check :biggrin:!

Anyone!?
 
The best check is to insert the result into the system described by equivalences picturing the same reality in a different way.
Notice that all the line of forces go through the center of the drum. If you relocate them there, the moment imposed on E won't change: the force is the same angle and magnitude, and the arm is also the same. Now let's compute the sum of the forces (The complex coordinate system is right to positive reals, up for positive imags, and I am just too addicted to express angles relative to due north.):
[tex] 120 - 60 \operatorname{sin}\left(-15\right) + 110 \operatorname{sin}\left(145\right) + \\<br /> 130 \operatorname{sin}\left(50.1\right) + 60 \mathbf{\imath} \operatorname{cos}\left(-15\right) + 110 \mathbf{\imath} \\<br /> \operatorname{cos}\left(145\right) + 130 \\<br /> \mathbf{\imath} \operatorname{cos}\left(50.1\right)[/tex]
unfortunately imag(F1+F2+F3+F4)*0.75= 37.3775097537 which is not quite the 50 I would expect.

Either I am or you are in error:)
 
I found a much simpler way to solve, I found that alpha = 34.27 degrees.
 
Last edited: