A 68.0 Kg tightrope walker stands at the center of a rope. The rope supports are 10 m apart and the rope sags 8.00 degrees at each end. The tightrope walker crouches down, then leaps straight up with an acceleration of 8.10 m/s^2 to catch a passing trapeze. What is the tension in the rope as he jumps?

Question

PbJ

Physics

25 January, 14:05

A 68.0 Kg tightrope walker stands at the center of a rope. The rope supports are 10 m apart and the rope sags 8.00 degrees at each end. The tightrope walker crouches down, then leaps straight up with an acceleration of 8.10 m/s^2 to catch a passing trapeze. What is the tension in the rope as he jumps?

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Answers (1)

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Alani Kemp · Answer 1

Tension = 4,373.70N

Explanation:

Ma = Sum of forces

Horizontal forces are:

0 = T1costheta=T2costheta

T1=T2=T

Vertical forces are:

Ma = T1sin8° + T2sin8° - mg

Ma = 2Tsin8° - mg

Therefore,

T = m (a+g) / (2sin8°)

T = 68 (8.10+9.8) / 2*0.139

68 (17.9) / 0.2783

T = 4373.697

T = 4373.70N