Posted by Simba on Friday, April 29, 2011 at 5:25pm.
first draw a position diagram, a horizontal line 10 m, and two sides of 5 m and 7 ml to get triangle ABC
where b = 10, a=7, and c=5
by the cosine law,
5^2 = 10^2 + 7^2 - 2(10)(7)cosC
cosC = 124/140
angle C = 27.66°
by the Sine Law
sinA/7 = sin27.66/5
angle A = 40.54
now draw a vector diagram, triangle PQR
where PR = 10 and a vertical line, P above R
PR represents the 10 kg mass
Using the above results,
angle P = 62.34
angle Q = 40.54+27.66 = 68.20
angle R = 49.46
by sine law:
PQ/sin 49.46 = 10/sin68.20
PQ = 8.18
QR/sin62.34 = 10/sin68.2
QR = 9.54
the tension is the 7 m string is 8.18 kg
the tension in the 5 m string is 9.54 kg
check my arithmetic
draw PQ parallel to BC and QR parallel to AB
angle QPR = 90-27.66 = 62.34°
the last two line got displaced, should have been:
first draw a position diagram, a horizontal line 10 m, and two sides of 5 m and 7 ml to get triangle ABC
where b = 10, a=7, and c=5
by the cosine law,
5^2 = 10^2 + 7^2 - 2(10)(7)cosC
cosC = 124/140
angle C = 27.66°
by the Sine Law
sinA/7 = sin27.66/5
angle A = 40.54
now draw a vector diagram, triangle PQR
where PR = 10 and a vertical line, P above R
PR represents the 10 kg mass
draw PQ parallel to BC and QR parallel to AB
angle QPR = 90-27.66 = 62.34°
Using the above results,
angle P = 62.34
angle Q = 40.54+27.66 = 68.20
angle R = 49.46
by sine law:
PQ/sin 49.46 = 10/sin68.20
PQ = 8.18
QR/sin62.34 = 10/sin68.2
QR = 9.54
the tension is the 7 m string is 8.18 kg
the tension in the 5 m string is 9.54 kg
check my arithmetic
Hey the guy above is totally right. Except for the fact that is screwed up the Lenght of PR. To get the right answer you have to do 10kg * 9.8N = 98N. Which is the length of PR. Now use 98 with sine law as Rimey did above and you shall get 80 and 94 for your two tensions.