91影视

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 $r=7.5m$

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.

$F_c=\frac{m{v}^2}{r}ormr{蝇}^2$

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

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

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

Determine the angular speed of seat

$$\begin{gathered} 蝇=\frac{2\mathrm{蟺苍}}{60} \\ =\frac{2\mathrm{蟺}脳28.0\mathrm{rpm}}{60} \\ =2.93rad/s \end{gathered}$$

Equating the vertical forces

$T_1\sin 50掳=mg$

Substituting all the values in above equation to get value of

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

Hence the tension in the upper cables is 1409.84N

Now equating the horizontal forces

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

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

$$\begin{gathered} 110.204kg脳7.5m脳{\left(2.93rad/s\right)}^2=T_2+1409.84N脳\cos 50掳 \\ {T}_2=7095.67N-906.22N \\ =6189.45N \end{gathered}$$

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