Missing picture: find it in attachment.
(a) 3214 N
To find the tension in the cable, we just need to analzye the forces along the vertical direction.
We have:
- The vertical component of the tension, which is [tex]T cos \theta[/tex], acting upward
- The weight of the passenger + the chair, mg, acting downward
Since there is equilibrium along this direction, we can write
[tex]T cos \theta - mg = 0[/tex]
where:
T is the magnitude of the tension
[tex]\theta=60^{\circ}[/tex]
m = 164 kg is the total mass of the chair and the occupant
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for T,
[tex]T=\frac{mg}{cos \theta}=\frac{(164)(9.8)}{cos 60}=3214 N[/tex]
(b) 12.2 m/s
To find the speed of the car, we have to analyze the forces in the horizontal direction.
The component of the tension in the cable provides the centripetal force that keeps the chair in circular motion, so:
[tex]T sin \theta = m \frac{v^2}{r}[/tex]
where
v is the speed
r is the radius of the circular path
We know that the length of the cable is
L = 10.1 m
So, the radius of the circle is
[tex]r=L sin \theta = (10.1)(sin 60)=8.7 m[/tex]
So now we can solve the equation to find v, the speed:
[tex]v=\sqrt{\frac{Tr sin \theta}{m}}=\sqrt{\frac{(3214)(8.7)(sin 60)}{164}}=12.2 m/s[/tex]
