Q 51E - University Physics with Modern Physics

Question

Question: In another version of the “Giant Swing” (see Exercise 5.50), the seat is connected to two cables, one of which is horizontal (Fig. E5.51). The seat swings in a horizontal circle at a rate of 28.0rpm. If the seat weighs 255N and an 825N person is sitting in it, find the tension in each cable.

Step-by-Step Solution

Answer

Tension in the upper cables is $1409.84 \text{N}$ and tension of in the horizontal cables is $6189.45 \text{N}$.

Step 1: Identification of given data

Weight of seat $W_c=255 \mathrm{N}$

Weight of person $W_p=255 \mathrm{N}$

Seat swing at a rate of $n=28.0 \mathrm{rpm}$

Radius of circular path $r=7.5 \text{m}$.

Step 2: Significance of centripetal force and centrifugal force

The force applied to an item in curved motion that is pointed toward the axis of rotation or the centre of curvature is known as a centripetal force.

A fictitious force that moves in a circle and is directed away from the centre of the circle is called centrifugal force.

$F_c=\frac{m{v}^2}{r} \text{or} mr{\omega }^2$

Where, m is mass of body, v is velocity of body, r is radius of circular path and the $\omega$ is angular speed.

Step 3: Determining the tension in each cable

Let assume that tension in upper cable is $T_1$ and the tension in the horizontal cable is $T_2$

Total weight of seat and person is 

 $\begin{array}{rcl}W& =& 255 \text{N}+825 \text{N}\\ & =& 1080\text{N}\end{array}$

Total mass of seat and person is 

 $\begin{array}{rcl}m& =& \frac{1080 \text{N}}{9.8 {\text{m/s}}^2}\\ & =& 110.204 \text{kg}\end{array}$

Determine the angular speed of seat

 $\begin{array}{rcl}\omega & =& \frac{2\pi n}{60}\\ & =& \frac{2\pi \times 28.0 \mathrm{rpm}}{60}\\ & =& 2.93 \text{rad/s}\end{array}$

Equating the vertical forces

 $T_1\sin {50}^0=mg$

Substituting all the values in above equation to get value of $\begin{array}{rcl}& & T_1\end{array}$

$\begin{array}{rcl}{T}_1& =& \frac{1080\text{N}}{\sin {50}^0}\\ & =& 1409.84 \text{N}\end{array}$

Hence the tension in the upper cables is $1409.84 \text{N}$

Now equating the horizontal forces

$$\begin{gathered} F_c=T_2+T_1\cos {50}^0 \\ mr{\omega }^2=T_2+T_1\cos {50}^0 \end{gathered}$$

Substituting all the values in above equation to the value of $T_2$

 $\begin{array}{rcl}110.204 \text{kg}\times \text{7.5} \text{m}\times {\left(\text{2.93} \text{rad/s}\right)}^2& =& T_2+1409.84 \text{N}\times \cos {50}^0\\ T_2& =& 7095.67 \text{N}-906.22 \text{N}\\ & =& 6189.45 \text{N}\end{array}$

Hence the tension in the horizontal cables is $6189.45 \text{N}$.