Question

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?

Answer 1

Answer:

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