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11111nata11111 [884]

2 years ago

8

In the Skycoaster amusement park ride, riders are suspended from a tower by a long cable. A second cable then lifts them until t

hey reach the starting position indicated in (Figure 1). The lifting cable is then released, and the riders swing down the circular arc shown.

2 answers:

Alborosie2 years ago

4 0

Answer:

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

inna [77]2 years ago

3 0

Answer:

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°

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