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Sedbober [7]
3 years ago
15

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.2 m/s .
What upward force must a branch provide to support the swinging gibbon?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
aleksklad [387]3 years ago
5 0

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force  T  is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force  in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T =  mg -  mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

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The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Given parameters

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To find

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Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.

In this case we have a vertical launch

          y = y₀ + v₀ t - ½ g t²

Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time

   

With the ball in hand, its position is zero

         0 = 0 + v₀ t - ½ g t²

         v₀ t - ½ g t² = 0

         v₀ = ½ g t

Let's calculate

         v₀ = ½ 9.8 2.4

         v₀ = 11.76 m / s

In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Learn more about vertical launch kinematics here:

brainly.com/question/15068914

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