Equilibrium of forces

Question:
When this system is in equilibrium, find the tension in cord b, $T_b$. The hanging mass is 2.00 kg.

Answer:
As shown in the second figure, the tensions are vectors on $x$-$y$ coordinate. Since these are at equilibrium, the summation of the vectors must be zero.

In order to calculate, we can use the component method.
 \[ T_{ax}=T_a \cos 270^o=0, T_{ay}=T_a \sin 270^o = -mg  \\
T_{bx}=T_b \cos 0^o=T_b, T_{by}=T_b \sin 0^o=0 \\
T_{cx}=T_c \cos 150^o, T_{cy}=T_c \sin 150^o  \]
Thus,
\[ T_{ax}+T_{bx}+T_{cx}=T_b+T_c\cos 150^o = 0 \\
   T_{ay}+T_{by}+T_{cy}=-mg+T_c\sin 150^o = 0\]
From the second equation, we have
\[ T_c = \frac{mg}{\sin 150^o}=39.2 \mathrm{N}\]
Then, plug it in the first equation.
\[ T_b = T_c\sin 150^o = 33.9 \mathrm{N}\]












Powered by Hirophysics.com