# Homework Help: Can anyone check my working on the tension upcoming test question

1. Nov 22, 2013

### cracktheegg

1. The problem statement, all variables and given/known data

[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]

2. Relevant equations

ƩMA = Ʃ r x F

3. 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

2. Nov 22, 2013

### tiny-tim

hi cracktheegg!

is your Rx positive or negative?

check the signs in your Ry

3. Nov 22, 2013

### PhanthomJay

Looks good to here, call it Ax
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.
Make correction above and this will make the magnitude of the reaction force at A, correct.

4. Nov 22, 2013

### cracktheegg

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?

5. Nov 22, 2013

### tiny-tim

hi cracktheegg!
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

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