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°
A 10kg mass is supported by two strings of length 5 m and 7 m attached to two points in the ceiling 10 m apart. Find the tension in each string.
Help will be much appreciated! :)
5 answers
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
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.
load of 10kg is hanged using two
strings 9m and 12m long tied to two point at the same
level 15m apart find the tension in the string
strings 9m and 12m long tied to two point at the same
level 15m apart find the tension in the string
This is not my question but it's like it so i will use it to try to do mine
1 answer