19.1: Mass Suspended by Two Ropes

As a typical example of a problem in statics, consider the situation shown in Fig. \(\PageIndex{1}\)(a). A block of mass \(m\) is suspended by a wire, and the upper end of the wire is attached to two more ropes or wires that connect to the ceiling. Each of the three ropes is under tension; the tensions are labeled \(T_{1}, T_{2}\), and \(T_{3}\).

To begin the analysis of this situation, it is often helpful to draw a free-body diagram for each body in the problem. A free-body diagram shows all the forces acting on the body, and helps clarify your thinking when doing the analysis. For this problem, there are two bodies present: the block and the knot. Fig. \(\PageIndex{1}\)(b) is a free-body diagram for the block, and Fig. \(\PageIndex{1}\)(c) is a free-body diagram for the knot.

Figure \(\PageIndex{1}\): A block suspended from the ceiling by ropes. (a) Diagram of the situation. The block of mass \(m\) is suspended by a rope; the upper end of the rope is attached to two other ropes that are attached to the ceiling. (b) Free-body diagram for the block. (c) Free-body diagram for the knot.

Now let's begin the analysis; our goal will be to determine the three tensions \(T_{1}, T_{2}\), and \(T_{3}\), given the mass \(m\) and two angles \(\theta_{1}\) and \(\theta_{2}\). First, let's look at the free-body diagram for the block (Fig. \(\PageIndex{1}\)(b)). For the block, the tension and weight vectors are given by

\[
\begin{align}
& \mathbf{T}_{3}=T_{3} \mathbf{j} \\[8pt]
& \mathbf{W}=-m g \mathbf{j}
\end{align}
\]

(Note the \(x\) and \(y\) directions indicated in Fig. \(\PageIndex{1}\)(a).) Now let's apply Newton's second law in both the \(x\) and \(y\) directions, noting that \(F=m a=0\) in this case:

\[
\begin{align}
x: \quad \sum F_{x}=m a_{x} & \Rightarrow & 0=0 \\[8pt]
y: \quad \sum F_{y}=m a_{y} & \Rightarrow & T_{3}-m g=0
\end{align}
\]

Both right-hand sides are zero because the acceleration of the block is zero. The \(x\) equation (Eq. \(\PageIndex{3}\)) yields a tautology \(0=0\), which gives us no information. The \(y\) equation (Eq. \(\PageIndex{4}\)) tells us \(T_{3}=m g\), so we've just found tension \(T_{3}\).

We can find the other two tensions ( \(T_{1}\) and \(T_{2}\) ) by analyzing the other body: the knot (Fig. \(\PageIndex{1}\)(c)). For the knot, the three tension vectors are given by

\[
\begin{align}
& \mathbf{T}_{1}=-T_{1} \cos \theta_{1} \mathbf{i}+T_{1} \sin \theta_{1} \mathbf{j} \\[8pt]
& \mathbf{T}_{2}=T_{2} \cos \theta_{2} \mathbf{i}+T_{2} \sin \theta_{2} \mathbf{j} \\[8pt]
& \mathbf{T}_{3}=-T_{3} \mathbf{j}
\end{align}
\]

Now let's apply Newton's second law ( \(F=m a=0\) ) individually to the \(x\) and \(y\) components:

\[
\begin{align}
x: \quad \sum F_{x}=m a_{x} & \Rightarrow & -T_{1} \cos \theta_{1}+T_{2} \cos \theta_{2}=0 \\[8pt]
y: \quad \sum F_{y}=m a_{y} & \Rightarrow & T_{1} \sin \theta_{1}+T_{2} \sin \theta_{2}-T_{3}=0
\end{align}
\]

Again both right-hand sides are zero because the knot is not accelerating. Since \(T_{3}\) is already known, this gives two simultaneous equations in the two unknown tensions \(T_{1}\) and \(T_{2}\). One method for solving this system of equations is to write the equations in matrix form:

\[
\left(\begin{array}{cc}
-\cos \theta_{1} \cos \theta_{2} \\[8pt]
\sin \theta_{1} \sin \theta_{2}
\end{array}\right)\left(\begin{array}{c}
T_{1} \\[8pt]
T_{2}
\end{array}\right)=\left(\begin{array}{c}
0 \\[8pt]
T_{3}
\end{array}\right)
\]

Now multiplying both sides on the left by the inverse of the \(2 \times 2\) matrix, we have

\[
\left(\begin{array}{c}
T_{1} \\[8pt]
T_{2}
\end{array}\right)=\left(\begin{array}{cc}
-\cos \theta_{1} \cos \theta_{2} \\[8pt]
\sin \theta_{1} \sin \theta_{2}
\end{array}\right)^{-1}\left(\begin{array}{c}
0 \\[8pt]
T_{3}
\end{array}\right)
\]

Since the tension \(T_{3}\) and the angles \(\theta_{1}\) and \(\theta_{2}\) are all known, this gives the two unknown tensions \(T_{1}\) and \(T_{2}\).

We can further simplify this by computing the matrix inverse explicitly. The determinant of the \(2 \times 2\) matrix is (Appendix 63.17)

\[\begin{align}
\operatorname{det}\left(\begin{array}{cc}
-\cos \theta_{1} \cos \theta_{2} \\[8pt]
\sin \theta_{1} \sin \theta_{2}
\end{array}\right) =-\cos \theta_{1} \sin \theta_{2}-\sin \theta_{1} \cos \theta_{2} \end{align} \]

\[ =-\sin \left(\theta_{1}+\theta_{2}\right)\]

and the matrix of cofactors is

\[
\operatorname{cof}\left(\begin{array}{cc}
-\cos \theta_{1} \cos \theta_{2} \\[8pt]
\sin \theta_{1} \sin \theta_{2}
\end{array}\right)=\left(\begin{array}{cc}
\sin \theta_{2} -\sin \theta_{1} \\[8pt]
-\cos \theta_{2} -\cos \theta_{1}
\end{array}\right)
\]

Hence the matrix inverse, which is the transposed matrix of cofactors divided by the determinant, is

\[
\begin{align}
\left(\begin{array}{cc}
-\cos \theta_{1} \cos \theta_{2} \\[14pt]
\sin \theta_{1} \sin \theta_{2}
\end{array}\right)^{-1} =-\frac{1}{\sin \left(\theta_{1}+\theta_{2}\right)}\left(\begin{array}{cc}
\sin \theta_{2} -\cos \theta_{2} \\[14pt]
-\sin \theta_{1} -\cos \theta_{1}
\end{array}\right) \end{align} \]

\[=\frac{1}{\sin \left(\theta_1+\theta_2\right)}\left(\begin{array}{cc}
-\sin \theta_2 & \cos \theta_2 \\[14pt]
\sin \theta_1 & \cos \theta_1
\end{array}\right) .
\]

The tensions \(T_1\) and \(T_2\) are therefore

\[
\begin{align}
\left(\begin{array}{l}
T_{1} \\
T_{2}
\end{array}\right) & =\frac{1}{\sin \left(\theta_{1}+\theta_{2}\right)}\left(\begin{array}{cc}
-\sin \theta_{2} & \cos \theta_{2} \\[14pt]
\sin \theta_{1} & \cos \theta_{1}
\end{array}\right)\left(\begin{array}{c}
0 \\
T_{3}
\end{array}\right) \\
& =\frac{T_{3}}{\sin \left(\theta_{1}+\theta_{2}\right)}\left(\begin{array}{c}
\cos \theta_{2} \\[14pt]
\cos \theta_{1}
\end{array}\right) .
\end{align}
\]

Recall that we've already found \(T_{3}=m g\); then the final results are

\[
\begin{align}
& T_{1}=\frac{m g \cos \theta_{2}}{\sin \left(\theta_{1}+\theta_{2}\right)}, \\[8pt]
& T_{2}=\frac{m g \cos \theta_{1}}{\sin \left(\theta_{1}+\theta_{2}\right)}, \\[8pt]
& T_{3}=m g .
\end{align}
\]