Skip to main content

General Physics I:

Section 4.2 Examples A

Example 4.2.1.

What is the tension in each of the strings of the system shown here:
Solution.
Step 1. List the given quantities and label the diagram:
Step 2. Isolate each object and draw the force diagrams:
Step 3. Resolve forces into components as shown in figure:
Step 4. Write equations using the figure
\begin{equation*} \sum{F_{x}} = T_{2}\cos\theta - T_{1}\sin\theta = 0 \end{equation*}
\begin{equation*} \sum{F_{y}} = T_{2}\sin\theta - T_{1}\cos\theta -T_{3}= 0 \end{equation*}
from figure
\begin{equation*} \sum{F_{y}} = T_{3}-w_{1}-T_{4}= 0 \end{equation*}
from figure
\begin{equation*} \sum{F_{y}} = T_{4}-w_{2} = 0 \end{equation*}
Step 5. Solve equations simultaneously for unknowns:
\begin{equation*} T_{4} = w_{2} = 10 \,N \end{equation*}
\begin{equation*} T_{3} = T_{4} + w_{1} = 30 \,N \end{equation*}
\begin{equation*} Now,\quad T_{2} = \frac{T_{1}\sin\theta}{\cos\theta}= T_{1}\tan\theta \end{equation*}
\begin{equation*} also,\quad T_{1} = \frac{T_{3}}{\tan\theta\sin\theta-\cos\theta} = 30 \,N; \end{equation*}
\begin{equation*} \therefore T_{2} =52 \,N \end{equation*}

Example 4.2.2.

Find the tension in each of the strings of the system below if the weight \(w=80 \,N\) and the angle \(\theta=30^{o}\text{.}\)
Solution.
From figure
\begin{equation*} \sum F_{x} =T_{2}\sin\theta-T_{1}\cos\theta =0 \end{equation*}
\begin{equation} \text{or,}\quad T_{1}=T_{2}\tan\theta =T_{2}\frac{1}{\sqrt{3}} \tag{4.2.1} \end{equation}
\begin{equation*} \sum F_{y} =T_{2}\cos\theta+T_{1}\sin\theta-T_{3} =0 \end{equation*}
\begin{equation*} \text{or,}\quad T_{2}\cos 30^{o}+T_{1}\sin 30^{o} =80 \end{equation*}
\begin{equation} \text{or,}\quad T_{2}\frac{\sqrt{3}}{2}+T_{1}\frac{1}{2} =T_{3} \tag{4.2.2} \end{equation}
\begin{equation*} \sum F_{y}=T_{3}-w=0 \end{equation*}
\begin{equation} T_{3}=w=80N \tag{4.2.3} \end{equation}
from eqns. (4.2.1), (4.2.2), and (4.2.3), we have
\begin{equation*} T_{2}=70 \,N \quad \text{and} \quad T_{1}=40 \,N \end{equation*}

Example 4.2.3.

A 5 kg block is held at rest against a vertical wall by a horizontal force of 100N.
  1. What is the frictional force exerted by the wall on the block?
  2. What is the minimum horizontal force needed to prevent the block from sliding down the wall if the coefficient of static friction between the wall and the block is \(0.4\text{?}\)
Solution.
Given: \(m= 5\,kg, \quad w= mg = 5\,kg\cdot9.8 \,m/s^{2} =49 \,N,\quad F=100 \,N.\)
  1. \begin{equation*} \sum F_{x}=F-N=0 \quad\Rightarrow\quad F=N=100\,N \end{equation*}
    \begin{equation*} \sum F_{y}=f-w=0\quad\Rightarrow\quad f=w=49\,N \end{equation*}
  2. \begin{equation*} f=\mu_{s}N \qquad\Rightarrow\quad N=\frac{f}{\mu_{s}}=\frac{49N}{0.4}=122\,N \end{equation*}

Example 4.2.4.

Draw force (free-body) diagrams for the bodies shown in the figure below:
Solution.