# Thread: Statics of particles in equilibirium

1. ## Statics of particles in equilibirium

Hello all, i am currently working on statics of particles, and am having problems on the following question.

What i done first was found the magnitude of the three angles using the cosine rule, i have both tried leaving the angles in exact form, and rounding off to the nearest decimal place, both giving me the wrong answer. The given angles are:
ABC = 42
BCA = 53
CAB = 85

What i do next was rearrange the vectors into a triangle of forces and apply the correct angles. I believe these are (after rearranging):
ABC = 127
BCA = 38
CAB = 15

maybe im wrong there, im not sure. After that i simply use the sine rule to find the tension in the two unknown strings, but i get the wrong answers >_<

T1 = 14.99 kg wt, T2 = 12.10 kg wt

Can anyone see what i am doing wrong?

2. Originally Posted by jrp131191
Hello all, i am currently working on statics of particles, and am having problems on the following question.

What i done first was found the magnitude of the three angles using the cosine rule, i have both tried leaving the angles in exact form, and rounding off to the nearest decimal place, both giving me the wrong answer. The given angles are:
ABC = 42
BCA = 53
CAB = 85

What i do next was rearrange the vectors into a triangle of forces and apply the correct angles. I believe these are (after rearranging):
ABC = 127
BCA = 38
CAB = 15

maybe im wrong there, im not sure. After that i simply use the sine rule to find the tension in the two unknown strings, but i get the wrong answers >_<

T1 = 14.99 kg wt, T2 = 12.10 kg wt

Can anyone see what i am doing wrong?

note that the problem is unorthodox since kg's are not a unit of weight, but I will go with that since the solutions are in kg's.

using the cosine law, angle between $T_1$ and $T_2$ is

$P = \arccos\left(\frac{10^2+12^2-15^2}{240}\right) \approx 85.46^\circ$

using the sine law, angle between $T_2$ and the horizontal is

$Q = \arcsin\left(\frac{10\sin{P}}{15}\right) \approx 41.65^\circ
$

angle between $T_1$ and the horizontal is

$R = 180 - P - Q \approx 52.89^\circ$

equilibrium in the x-direction ...

$T_1\cos{R} - T_2\cos{Q} = 0$

equilibrium in the y-direction ..

$T_1\sin{R} + T_2\sin{Q} = 20$

solve the system for $T_1$ and $T_2$ and you should get the desired results.