Can anyone check my working on the tension upcoming test question

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Homework Help Overview

The discussion revolves around a physics problem involving tension in a system, likely related to forces acting on a pulley or beam setup. Participants are analyzing calculations related to tension, reaction forces, and equilibrium conditions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are reviewing calculations for tension and reaction forces, questioning the signs and components of forces in the y-direction. Some suggest re-evaluating the approach to finding resultant forces, while others express confusion about the necessity of certain calculations.

Discussion Status

There are multiple interpretations of the problem, with participants providing feedback on each other's calculations. Some guidance has been offered regarding the signs and components of forces, indicating a productive exchange of ideas. However, no consensus has been reached on the best approach to the problem.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available for solving the problem. There are indications of potential errors in calculations and assumptions that are being discussed.

cracktheegg
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Homework Statement



[URL=http://s1345.photobucket.com/user/Duk_Bato/media/Untitled_zps84f60bfd.png.html][PLAIN]http://i1345.photobucket.com/albums/p679/Duk_Bato/Untitled_zps84f60bfd.png[/URL][/PLAIN]

Homework Equations



ƩMA = Ʃ r x F

The Attempt at a Solution



1.6 * T+ 0.8 * sin 40 *T = 1.2 * 250
T (1.6 + 0.8 * sin 40 ) = 300
T = 300 / ( 1.6 + 0.8 * sin 40 )
T = 141.9 N

RX= 108.701

FY=141.9+141.9+250=533.8

RF= Root(533.8^2 + 108.701^2) = 544.76N
 
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hi cracktheegg! :smile:

your T looks ok

is your Rx positive or negative?

check the signs in your Ry :wink:
 
cracktheegg said:

Homework Statement



[URL=http://s1345.photobucket.com/user/Duk_Bato/media/Untitled_zps84f60bfd.png.html][PLAIN]http://i1345.photobucket.com/albums/p679/Duk_Bato/Untitled_zps84f60bfd.png[/URL][/PLAIN]

Homework Equations



ƩMA = Ʃ r x F

The Attempt at a Solution



1.6 * T+ 0.8 * sin 40 *T = 1.2 * 250
T (1.6 + 0.8 * sin 40 ) = 300
T = 300 / ( 1.6 + 0.8 * sin 40 )
T = 141.9 N

RX= 108.701
Looks good to here, call it Ax
FY=141.9+141.9+250=533.8
You have some errors here. You are looking for Ay by summing forces in the y direction = 0. You have signage errors and you forgot to calculate the vertical component of one of your tension forces. Assume Ay acts up, and try summing forces in the y direction = 0 again.
RF= Root(533.8^2 + 108.701^2) = 544.76N
Make correction above and this will make the magnitude of the reaction force at A, correct.
 
I find the pulley Resultant force= 257.21

0=141.9cos40 - Rdy (up)
0=-141.9sin40 - 141.9 +Rdx (right)
root(Rdy^2+rdx^2)= 257.21N

tan^-1(Rdy/rdx) =65

-250+257.21sin65+Ray=0
Ray=16(up)

RX= 108.701

Rf= root(108.701^2+16^2)= correct?
 
hi cracktheegg! :smile:
cracktheegg said:
0=141.9cos40 - Rdy (up)
0=-141.9sin40 - 141.9 +Rdx (right)
root(Rdy^2+rdx^2)= 257.21N

i think you have your Rdx and Rdy the wrong way round (though that won't affect your final result)

but i don't understand why you're doing it this way …

the question doesn't ask for the pulley resultant, and finding it takes much longer and gives more opportunity for making mistakes :redface:

why didn't you simply do Ry for the beam?
 

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