Asked by Mat
Need a bit of help solving the following equations.
sqrt(5/8-(sqrt(5)/8))LH+cos (43pi/180)RH = 0
and
1/4(1+sqrt(5)LH+sin(43pi/180)RH-2g=0
What is the best way to start when trying to combine these equations?
Thanks
sqrt(5/8-(sqrt(5)/8))LH+cos (43pi/180)RH = 0
and
1/4(1+sqrt(5)LH+sin(43pi/180)RH-2g=0
What is the best way to start when trying to combine these equations?
Thanks
Answers
Answered by
Steve
what do the LH and RH mean?
Answered by
Mat
Just my variables, its to find tension in two strings. I made them LH and RH in my drawings.
I could have easily put T1 and T2 come to think of it...
I could have easily put T1 and T2 come to think of it...
Answered by
Steve
I'll just use x and y, giving
√(5/8-(√5/8))x+cos(43°)y = 0
0.58778x + 0.73135y = 0
1/4(1+√5)x+sin(43°)y-2g=0
0.80902x + 0.68200y = 19.6
Now just solve by your favorite method (I suggest substitution).
√(5/8-(√5/8))x+cos(43°)y = 0
0.58778x + 0.73135y = 0
1/4(1+√5)x+sin(43°)y-2g=0
0.80902x + 0.68200y = 19.6
Now just solve by your favorite method (I suggest substitution).
Answered by
Mat
Thanks steve, out of interest ive gotten a negative value for one. Since im looking for the tension in L and R strings. That doesnt seem right in which case ive made a mistake in getting to this point.
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