Can anyone check my working on the tension upcoming test question

AI Thread Summary
The discussion revolves around checking calculations related to tension and forces in a physics problem involving a pulley system. The initial calculations for tension (T) and horizontal force (RX) appear correct, but there are noted errors in the vertical force (FY) calculations, particularly regarding signage and the vertical component of tension. Participants suggest re-evaluating the forces in the y-direction and correcting the approach to simplify the problem. There is also a debate about the necessity of calculating the pulley resultant force, with some arguing it complicates the solution unnecessarily. Overall, the focus is on ensuring accurate calculations and simplifying the problem-solving process.
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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