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A. A 0.60 kg bullfrog is sitting at rest on a level log. How large is the normal force of the log on the frog?

b. A second 0.60 kg bullfrog is on a log tilted 0.30° above horizontal. How large is the normal force of the log on this frog?

User Nandun
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1 Answer

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Answer:

a. 5.88 N

b. 5.09 N

Step-by-step explanation:

Check attachment for drawings of the free-body diagrams.

(a) As we can see on our free-body diagram, the weight force and the normal force are the only forces acting on the y-direction.

Since the bullfrog is in a state of equilibrium, due to it being at rest, we can say that the weight force and normal force are equal to each other.

The weight force w is equal to mg, so we can plug the weight of the frog (0.60 kg) and the force of gravity (9.8 m/s²) to find the weight force.

  • w = mg
  • w = (0.60 kg)(9.8 m/s²)
  • w = 5.88 N

The normal force and the weight force are equal to each other; therefore, the normal force is also equal to 5.88 N.

(b) As seen on the free-body diagram, the two forces that are in the y-direction are the normal force and the y-component of the weight force.

Therefore, by finding the force of the y-component of the weight force, we can find the normal force since they are equal to each other because the bullfrog is not moving up or down.

  • F_y = n - w cos(30) = 0
  • F_y = n = w cos(30)

We know that the weight force is equal to 5.88 N, so we can plug this value into the y-component for the weight force:

  • w cos(30)
  • 5.88 * cos(30) = 5.09 N

Since the y-component of the weight force and the normal force are equal to each other, the normal force is also equal to 5.09 N.

A. A 0.60 kg bullfrog is sitting at rest on a level log. How large is the normal force-example-1
User Premal Khetani
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