Original post by DFranklin
Use the 1st equation to write Tq in terms of Tp.
Tp = Tq Coss35 / cos55?
Original post by r7kytt
Tp = Tq Coss35 / cos55?
yes exactly.
Now input this Tp value (Tq cos35/ cos 55) into equation 2's Tp.
Original post by qwert7890
yes exactly.
Now input this Tp value (Tq cos35/ cos 55) into equation 2's Tp.
Now input this Tp value (Tq cos35/ cos 55) into equation 2's Tp.
I got (Tqcos35 ) + (Tqsin35sin55) / sin55 = 2g
Original post by r7kytt
I got (Tqcos35 ) + (Tqsin35sin55) / sin55 = 2g
How?
Tp = (Tq Cos 35) / cos 55
Insert this into equation 2's Tp
Equation 2 --> (Tp sin 55) + (Tq sin 35) = 2g
Replacing Tp you get
--> [ [ (Tq Cos 35) / (cos 55) ] x sin 55 ] + (Tq sin 35) = 2g
Solve further and you get:
[ (Tq cos 35) x (tan 55) ] +
Each bold, italic and underlined part corresponds to the same part.
(edited 3 years ago)
Original post by qwert7890
How?
Tp = (Tq Cos 35) / cos 55
Insert this into equation 2's Tp
Equation 2 --> (Tp sin 55) + (Tq sin 35) = 2g
Replacing Tp you get
--> [ [ (Tq Cos 35) / (cos 55) ] x sin 55 ] + (Tq sin 35) = 2g
Solve further and you get:
[ (Tq cos 35) x (tan 55) ] +
Each bold, italic and underlined part corresponds to the same part.
Tp = (Tq Cos 35) / cos 55
Insert this into equation 2's Tp
Equation 2 --> (Tp sin 55) + (Tq sin 35) = 2g
Replacing Tp you get
--> [ [ (Tq Cos 35) / (cos 55) ] x sin 55 ] + (Tq sin 35) = 2g
Solve further and you get:
[ (Tq cos 35) x (tan 55) ] +
Each bold, italic and underlined part corresponds to the same part.
Yeah, but don’t you make denominator the same?
Original post by r7kytt
Yeah, but don’t you make denominator the same?
The only denominator would go away if you do sin/cos = tan.
Original post by qwert7890
The only denominator would go away if you do sin/cos = tan.
So from here, what do I do? Times both sides by sin55?
Original post by r7kytt
So from here, what do I do? Times both sides by sin55?
Can you post your entire working?
If you could post a picture instead of typing I'd really appreciate it! (Perhaps just write it on a piece of paper and upload?)
Original post by r7kytt
So from here, what do I do? Times both sides by sin55?
You seem to be making this unnecessarily complex. This is just GCSE simultaneous equations with different numbers
You already know that where k = (cos 35 / cos 55)
Putting this into the 2nd equation gives you
or
Therefore
Now just plug in the numbers on your calculator. Taking g = 9.8 I get T_q = 11.242 to 3d.p. or approx 11.2 as required.
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