Solving Tension Problem: Can't Find T1

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The discussion revolves around solving for the tension T1 in a physics problem involving forces at knots. The user initially calculated T1 incorrectly and was advised to apply Newton's first law in both the x and y directions to set up equations for T1 and T2. The correct equations were established as T1sin55 - T2sin10 = 90N for vertical forces and T1cos55 - T2cos10 = 0 for horizontal forces. After resolving the equations, the user found T1 to be approximately 125.35 N and subsequently calculated T2 and other tensions. The final values for T3 and weight W were also discussed, confirming the calculations with slight variations among participants.
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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
 
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i have to start solving all of the problems contained by first gettin T1 value correct?
 
anglum said:
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
 
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 teh top to the knot ... then knew the force of that had to be 90 N... thus sin55 = 90/T1 correctt?
 
if i solve for T1 that way i get 109.8678N and that is incorrect
 
anglum said:
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 teh 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.
 
so my equation to solve looks like this

T1sin55 + T2sin10 = 90N ?
 
anglum said:
so my equation to solve looks like this

T1sin55 + T2sin10 = 90N ?
Watch your signs. T1 component acts up, T2 component acts down.
 
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?
 
  • #10
anglum said:
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.
 
  • #11
how am i supposed to solve for T1 and T2 ? combine those equations?
 
  • #12
anglum said:
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.
 
  • #13
if i solve for T2 in the 2nd equation i get -T2 = -T1cos55/cos10
 
  • #14
anglum said:
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...
 
  • #15
ok so then i get where T1 = X

.81915X - .57357X/.98480 = 90 ?
 
  • #16
anglum said:
ok so then i get where T1 = X

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

you forgot to multiply by sin10
 
  • #17
o sooo

.8195X - .57357X/.98480 (.173648) = 90?
 
  • #18
looks right.
 
  • #19
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?
 
  • #20
my math has to be wayyyy off
 
  • #21
anglum said:
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
 
  • #22
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
 
  • #23
but once i have T1 and T2 how do i get T3 and the other weight?
 
  • #24
anglum said:
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.
 
  • #25
ok the horizontal force on the 2nd knot is ---- cos10 * 73.004 correct?
 
  • #26
anglum said:
ok the horizontal force on the 2nd knot is ---- cos10 * 73.004 correct?

what about T3?
 
  • #27
and the vertical force on it is

sin10(73.004) + W + T3sin43 = 0

?
 
  • #28
the horizontal force is cos10(73.004) + T3cos43 = 0 ??
 
  • #29
anglum said:
the horizontal force is cos10(73.004) + T3cos43 = 0


??
watch your plus and minus signs, and note the angle is 43 degrees, correct, sorry.
 
Last edited:
  • #30
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?
 
  • #31
anglum said:
and the vertical force on it is

sin10(73.004) + W + T3sin43 = 0

?
Yes, except watch plus and minus signs.
 
  • #32
how should the plus minus signs be? and isn't the angle 43? for horizontal and vertical
 
  • #33
ok I am assuming this is what they should look like

cos10(73.004) - T3cos43 = 0 HORIZONTAL

sin10(73.007) + T3sin43 - W = 0 VERTICAL

did i fix them
 
  • #34
Yeah, those look right.
 
  • #35
so i solve for T3 in the horizontal and get 98.303N

then plug that into the vertical and get W = 92.3971N or 9.428275kg

are those the answers you got?
 
  • #36
anglum said:
so i solve for T3 in the horizontal and get 98.303N

then plug that into the vertical and get W = 92.3971N or 9.428275kg

are those the answers you got?

I get something different for W. But I get the same T3.
 
  • #37
for W did u use the equation

sin10(73.007) + T3sin43 - W = 0

12.677 + (98.303)(.681998) - W = 0

12.677 + 67.042 = W

W = 79.719484 N or 8.13464 kg?
 
  • #38
anglum said:
for W did u use the equation

sin10(73.007) + T3sin43 - W = 0

12.677 + (98.303)(.681998) - W = 0

12.677 + 67.042 = W

W = 79.719484 N or 8.13464 kg?

yeah that looks right.
 
  • #39
thanks again
 
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