Solving Tension Problem: Can't Find T1

[anglum]

since there is a picture for this problem i am providing a link to view the problem

http://i199.photobucket.com/albums/aa314/anglum/help2.jpg


i solved for tension of T1 by taking the sin55=90/T1
solved for T1 and converted to kg and got 11.211 kg which was incorrect

since the remainder of the problems are dependent on having T1 right i am now stuck

not sure what to do since the forces on the "knots" is equal to zero i used pythagorean theorem to get my answer since i knew the vertical force on T1 was 90N

please help

thank you

 

[anglum]

i have to start solving all of the problems contained by first gettin T1 value correct?

 

[PhanthomJay]

since there is a picture for this problem i am providing a link to view the problem

http://i199.photobucket.com/albums/aa314/anglum/help2.jpg


i solved for tension of T1 by taking the sin55=90/T1

what about the T2 force at this knot? You've got to use Newton 1 in both the x and y directions to solve for the 2 unknown forces with the 2 equations.

solved for T1 and converted to kg and got 11.211 kg which was incorrect

why are you converting a force to a mass unit?

since the remainder of the problems are dependent on having T1 right i am now stuck

not sure what to do since the forces on the "knots" is equal to zero i used pythagorean theorem to get my answer since i knew the vertical force on T1 was 90N

redo the T1 calc

 

[anglum]

i am converting the force to Kg because part 1 of the problem asks for tension in T1 in units of kg

i drew a vertical line down from the top to the knot ... then knew the force of that had to be 90 N... thus sin55 = 90/T1 correctt?

 

[anglum]

if i solve for T1 that way i get 109.8678N and that is incorrect

 

[PhanthomJay]

i am converting the force to Kg because part 1 of the problem asks for tension in T1 in units of kg

i drew a vertical line down from the top to the knot ... then knew the force of that had to be 90 N... thus sin55 = 90/T1 correctt?

No. You must isolate the knot and note that there is both a T1sin 55 and a T2sin 10 componnent in the vertical direction, one acting up and the other down, the sum total of which algebraically adds to 90 Newtons. Look in the x direction as well and apply Newton 1 again. You get 2 equations with 2 unknowns, which you can now solve for T1 and T2. I don't know why you would convert the tension to kilos, must be a misprint.

 

[anglum]

so my equation to solve looks like this

T1sin55 + T2sin10 = 90N ?

 

[PhanthomJay]

so my equation to solve looks like this

T1sin55 + T2sin10 = 90N ?

Watch your signs. T1 component acts up, T2 component acts down.

 

[anglum]

ok so itd be T1sin55 - T2sin10 = 90N? and that is just the vertical tension on that one

and the horizontal would be T1cos55 - T2cos10 = 0?

 

[PhanthomJay]

ok so itd be T1sin55 - T2sin10 = 90N? and that is just the vertical tension on that one

and the horizontal would be T1cos55 - T2cos10 = 0?

Yes, good, solve for T1 and T2, then move to the next knot.

 

[anglum]

how am i supposed to solve for T1 and T2 ? combine those equations?

 

[learningphysics]

ok so itd be T1sin55 - T2sin10 = 90N? and that is just the vertical tension on that one

and the horizontal would be T1cos55 - T2cos10 = 0?

Yes, the equations are right. Solve for T2 in one equation and plug it into the other equation.

 

[anglum]

if i solve for T2 in the 2nd equation i get -T2 = -T1cos55/cos10

 

[learningphysics]

if i solve for T2 in the 2nd equation i get -T2 = -T1cos55/cos10

yup, so T2 = T1cos55/cos10, you can plug in cos55 and cos10...

 

[anglum]

ok so then i get where T1 = X

.81915X - .57357X/.98480 = 90 ?

 

[learningphysics]

ok so then i get where T1 = X

.81915X - .57357X/.98480 = 90 ?

you forgot to multiply by sin10

 

[anglum]

o sooo

.8195X - .57357X/.98480 (.173648) = 90?

 

[learningphysics]

looks right.

 

[anglum]

so i then get .8159X -.57357x/.98480 = 90/.173648

then i get

.8159X - .57357X = (90/.173648) * (.98480)

.24233X = 510.4118677

T1 = 2106.267766 that can't be right?

 

[anglum]

my math has to be wayyyy off

 

[learningphysics]

so i then get .8159X -.57357x/.98480 = 90/.173648

nope... this isn't right.

the equation is:

.81915X - .57357X/.98480 = 90

Not this:

(.81915X - .57357X)/.98480 = 90

 

[anglum]

yeah I am an idiot... when i solve that the right way i get T1 = 125.3455633 N

then i can plug that in and solve for T2 and get T2 = 73.004N

 

[anglum]

but once i have T1 and T2 how do i get T3 and the other weight?

 

[learningphysics]

but once i have T1 and T2 how do i get T3 and the other weight?

do the freebody diagram equations of the second knot.

 

[anglum]

ok the horizontal force on the 2nd knot is ---- cos10 * 73.004 correct?

 

[learningphysics]

ok the horizontal force on the 2nd knot is ---- cos10 * 73.004 correct?

what about T3?

 

[anglum]

and the vertical force on it is

sin10(73.004) + W + T3sin43 = 0

?

 

[anglum]

the horizontal force is cos10(73.004) + T3cos43 = 0 ??

 

[PhanthomJay]

the horizontal force is cos10(73.004) + T3cos43 = 0


??

watch your plus and minus signs, and note the angle is 43 degrees, correct, sorry.

 

[anglum]

what should the plus minus signs be?

do i have the equations for vertical and horizontal right?

i thought the angle on the problem was 43 degrees?