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An 80-kg clown sits on a 20-kg bike on a tightrope attached between two trees. The center of mass of the clown is 1.6 m above the rope, and the center of mass of the bike is 0.70 m above the rope.

A) A load of what mass should be fixed onto the bike and hang 2.10 m below the rope so that the center of mass of the clown-bike-load system is 0.50 m below the rope?

B) What is the force that the rope exerts on each tree if the angle between the rope and the horizontal is 15 (degrees)?

Note: I have tried the center of mass formula Xcm= [(m1*x1) + (m2*x2) + (m3*x3)]/(m1+m2+m3) with no luck. Any help is much appreciated.

Answer :

Answer:

Explanation:

the center of mass formula

Ycm= [(m₁y₁) + (m₂y₂) + (m₃y₃)] / (m₁+m₂+m₃)

Rope forms the x axis and position of centre of different massses are above or below it so they represent their location on y - axis.

y₁ = 1.6 , y₂ = .7 and y₃ = - 2.1

Ycm ( given ) = - .5

Putting the values of masses and positions

- .5 = 80 x 1.6 + 20 x .7 + m₃ x - 2.1 / ( 80 + 20 + m₃ )

- .5 = 128  + 14  + m₃ x - 2.1 / ( 100+ m₃ )

- 50 - .5 m₃ = 142 - 2.1 m₃

1.6 m₃ = 192

m₃ = 120 kg .

B )

Total downward force is weight of  total mass  = 80 + 20 + 120

= 220 kg

weight = 220  x 9.8 = 2156 N .

component of weight perpendicular to rope

= 2156 cos 15 = 2082.53 N

This force will be equally distributed over each tree , so force on each tree =  2082.53 / 2 = 1041.26 N .

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