Q51E - University Physics with Modern Physics

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 . If the seat weighs and an person is sitting in it, find the tension in each cable.

                                               

Step-by-Step Solution

Answer

Tension in the upper cables is 1409.84 N and tension of in the horizontal cables is 6189.45 N

Step 1: Identification of given data

Weight of seat $W_c=255N$

Weight of person $W_p=255N$

Seat swing at a rate of  $n=28.0rpm$

Radius of circular path  $r=7.5m$

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.

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

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 

$$\begin{gathered} W=255N+825N \\ =1080N \end{gathered}$$ 

Total mass of seat and person is 

$$\begin{gathered} W=\frac{1080N}{9.8 m/s^2} \\ =110.204kg \end{gathered}$$ 

Determine the angular speed of seat

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

Equating the vertical forces

$T_1\sin 50°=mg$ 

Substituting all the values in above equation to get value of

$$\begin{gathered} T_1=\frac{1080N}{\sin 50°} \\ =1409.84N \end{gathered}$$

Hence the tension in the upper cables is 1409.84N

Now equating the horizontal forces

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

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

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

Hence the tension in the horizontal cables is $6189.45N$