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Chapter 7: Q. 48 (page 180)

A 70kg tightrope walker stands at the center of a rope. The rope supports are 10m apart and the rope sags 10掳 at each end. The tightrope walker crouches down, then leaps straight up with an acceleration of 8.0m/s2 to catch a passing trapeze. What is the tension in the rope as he jumps?

Short Answer

Expert verified

The tension in the rope as he jump is 3588 N.

Step by step solution

01

Net force.

Net force:

The resultant of all the forces acting on an object is called the net force. T

Lets find the net force by

Fnet=n-FG=ma

02

Estimating net force.

Calculate Net force

n=ma+FG

=m(a+g)

=ma+mg

=(70kg)(8.0m/s2+9.8m/s2)

=1246N

03

Tightrope.

There are two tension forces acting 10o upward on the tightrope,

As the same force acting downward on the person.

04

y-direction.

The resultant force in y direction is zero.

Which is given as .

Fnety=2Tsin10-n=0

05

Solving the  tension force T

Calculate Tension as below

T=n2(sin10o)

Substitute the value we get,

=1246N2sin10

=3588N

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