Cat on a Plank Torque Problem: How Close Can the Cat Walk Before the Plank Tips?

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In summary, a cat with a mass of 3.3 walks along a 4.20 meter long plank with a mass of 5.00 kg, supported by two sawhorses. The plank is just beginning to tip when the cat is 0.59 meters from the right end of the plank. This is found by equating the clockwise torque due to the cat's weight with the counter-clockwise torque due to the plank's weight.
  • #1
Andrews989
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Homework Statement



A cat walks along a uniform plank that is 4.20 long and has a mass of 5.00 . The plank is supported by two sawhorses, one 0.440 from the left end of the board and the other 1.50 from its right end.

If the cat has a mass of 3.3 , how close to the right end of the two-by-four can it walk before the board begins to tip?

5416311018_42.jpg


Homework Equations



Net force = 0
Net torque = 0

F=ma

Torque = r * F * sin(theta)

The Attempt at a Solution



Since the board is just beginning to tip, there is no weight on the left sawhorse.

So the right side the torque is equal to the force (mass * g) times the distance which is the unkown?

Force = 3.3*9.81 = 32.37 N

Torque = 32.37 N * d?

I'm stuck...
 
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  • #2
So far, so good. You found an expression for the clockwise torque due to the cat's weight. What's providing the counter-clockwise torque?
 
  • #3
Doc Al said:
So far, so good. You found an expression for the clockwise torque due to the cat's weight. What's providing the counter-clockwise torque?

Mass of the board. Which you count the mass from the center of gravity.

So, the plank is 4.2m. The center is 2.1m. The axis of rotation is 1.5m from the right. Making the plank .6 meters from the axis of rotation (to the left).

The the torque of the plank is .6m times 5kg * 9.81 = 29.3.So the torque on the left has to equal the torque on the right?

29.3 = 32.37 N * d

d = .91 meters?

It's online homework and that is wrong... so something is wrong...
 
  • #4
You found the cat's distance from the pivot. How far is that from the end?
 
  • #5
Doc Al said:
You found the cat's distance from the pivot. How far is that from the end?

Oh. Didn't think about that. So 1.5 - .91 = .59 meters...which is the correct answer. Thanks!
 

1. What is the "Cat on a Plank Torque Problem"?

The "Cat on a Plank Torque Problem" is a thought experiment that demonstrates the concept of torque in physics. It involves a cat standing on a plank that is balanced on a pivot point, and explores how the cat's weight and position on the plank affect the overall torque of the system.

2. Why is this problem important in science?

This problem is important because it allows us to understand the relationship between force, distance, and rotation in a real-life scenario. It also has practical applications in engineering and mechanics, such as in designing structures and machines that need to balance or rotate.

3. What factors affect the torque in this problem?

The torque in this problem is affected by the weight of the cat, the length of the plank, the distance of the cat from the pivot point, and the angle at which the plank is positioned. All of these factors contribute to the overall torque and can be manipulated to change the outcome of the experiment.

4. How does torque relate to other concepts in physics?

Torque is closely related to the concepts of force, distance, and rotation. It is also connected to angular acceleration, which is the rate of change of angular velocity. In addition, torque is a crucial component in understanding rotational motion and equilibrium.

5. What are some real-world examples of torque in action?

There are many real-world examples of torque, such as using a wrench to turn a bolt, a door being opened or closed, a balance scale, and even the rotation of the Earth. In sports, torque is also important in activities such as throwing a ball or swinging a bat or golf club. In everyday life, torque is present in tasks like using a can opener or turning a key in a lock.

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