In the Skycoaster amusement park ride, riders are suspended from a tower by a long cable. A second cable then lifts them until they reach the starting position indicated in (Figure 1). The lifting cable is then released, and the riders swing down the circular arc shown.

T1 cos 15 = T2 cos 26

T1 = T2 cos 26 / cos 15

T1 = 0.930 T2 (1)

T1 sin 15 + T2 sin 26 = m g

substitute eqn(1)

(0.930 T2 ) sin 15 + T2 sin 26 = 230 * 9.8

0.241 T2 + 0.438 T2 = 2254

Tension in right cable T2 = 3319 N

substitute in eqn (1)

T1 = 0.930 * 3319

Tension in left cable T1 = 3087

T1 cos 15 = T2 cos 26

T1 = T2 cos 26 / cos 15

T1 = 0.930 T2 (1)

T1 sin 15 + T2 sin 26 = m g

substitute eqn(1)

(0.930 T2 ) sin 15 + T2 sin 26 = 230 * 9.8

0.241 T2 + 0.438 T2 = 2254

Tension in right cable T2 = 3319 N

substitute in eqn (1)

T1 = 0.930 * 3319

Tension in left cable T1 = 3087 N

Explanation:

If the four riders have a total mass of 230 kg, what is the tension in the left cable just before release?

Express your answer with the appropriate units.

15° 26°

real, smaller and upside-down.   so the answer should be b.

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