Question:

When this system is at equilibrium, find the tension in cord b, \(T_b\), knowing that the hanging mass is 2.00 kg.

Answer:

As shown in the figure below, the tensions are vectors on the \(x\)-\(y\) coordinate. Since these are at equilibrium, the sum of the vectors must be zero.

In order to calculate, we can use the component method, so each x- and y-components are expressed as follows:

\[ T_{ax}=T_a \cos 270^o=0, T_{ay}=T_a \sin 270^o = -mg  \]

\[ T_{bx}=T_b \cos 0^o=T_b, T_{by}=T_b \sin 0^o=0 \]

\[ T_{cx}=T_c \cos 150^o, T_{cy}=T_c \sin 150^o  \]

Thus, we can add all of x’s and y’s.

\[ T_{ax}+T_{bx}+T_{cx}=T_b+T_c\cos 150^o = 0 \]

\[   T_{ay}+T_{by}+T_{cy}=-mg+T_c\sin 150^o = 0\]

From the second equation, we have

\[ T_c = \frac{mg}{\sin 150^o}=39.2 \mathrm{N}\]

Then, plug it in the first equation.

\[ T_b = T_c\sin 150^o = 33.9 \mathrm{N}\] This is the answer!