Posted by **Simba** on Friday, April 29, 2011 at 5:25pm.

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! :)

- Calculus/ Vectors -
**Reiny**, Friday, April 29, 2011 at 5:47pm
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°

- retyped - Calculus/ Vectors -
**Reiny**, Friday, April 29, 2011 at 5:50pm
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

- Calculus/ Vectors -
**Sam**, Sunday, May 1, 2011 at 4:16pm
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.

## Answer This Question

## Related Questions

- Vectors - A 10-kg mass is suspended by two strings of length 5m and 7m attached ...
- Vectors - A 10-kg mass is suspended by two strings of length 5m and 7m attached ...
- physics - a 10-kg mass is supported by two strings of length 5m and 7m attached ...
- physics - A doubly tethered model airplane of mass 1.0 kg is attached to a ...
- Physics - A 1.34‒kg ball is attached to a rigid vertical rod by means of ...
- Vectors - A mass 0f 10kg is suspended by 2 pieces of string, 30 cm and 40 cm ...
- Mechanics (A level), Please help - A light string of length a is attached to two...
- physics - A 0.101 kg meter stick is supported at its 40 cm mark by a string ...
- physics - the ends of a light string are tied to two hooks A and Bin are ceiling...
- physics - A 0.105 kg meter stick is supported at its 40 cm mark by a string ...

More Related Questions