Static Equilibrium Question

In summary, the problem is about rescuing a massless cat stuck at the end of a 210lb telephone pole that is sticking out over the Grand Canyon, with a total length of 90 feet and 32 feet sticking out. The person trying to rescue the cat weighs 190lbs and needs to know how far out they can walk onto the pole to rescue the cat. The solution involves understanding torque and balancing the torque of the person with the torque of the ground section of the pole. However, it is mentioned that the cat being massless means it will not fall, so risking one's life may not be necessary.
  • #1
Planckenstein
4
0
I've a test soon, and I've been thinking about this, and not quite sure how to approach it. I think I understand the chapter, but I can't seem to get it.

A uniform 210lb telephone pole is sticking out over the Grand Canyon (32 feet of it is sticking out) with a massless cat stuck at the end of it. If you weigh 190lbs, how far out can you walk onto the pole to rescue the cat.

This seems relatively easy but, again, I just can't get it. Answer's suppost to be 14.4ft. Thanks!
 
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  • #2
Planckenstein said:
A uniform 210lb telephone pole is sticking out over the Grand Canyon (32 feet of it is sticking out) with a massless cat stuck at the end of it. If you weigh 190lbs, how far out can you walk onto the pole to rescue the cat.
You will have to give us all the information. I gather the pole has a total length greater than 64 feet. How much longer than 64 feet is it?

AM
 
  • #3
Whoops! The pole is 90 feet long. That's all the information I've been given.
 
  • #4
Planckenstein said:
Whoops! The pole is 90 feet long. That's all the information I've been given.

This is a torque problem about a fulcrum at the cliff edge. Think of the 58 feet of pole on the ground as a weight of 58/90ths of the pole weight with a centre of mass located at the midpoint of that 58 foot length, or 29 feet from the fulcrum. Similarly the 32 foot section is a mass of 32/90ths of 210 with a centre of mass at 16 feet. Add the torque caused by you at a distance s from the fulcrum and that has to equal the torque of the ground section.

But why risk your life? If it is a massless cat, it is not going to fall!

AM
 

1. What is static equilibrium?

Static equilibrium is a state in which an object is at rest and has no acceleration. This means that the net force acting on the object is zero, and the object will remain in this state unless acted upon by an external force.

2. What are the conditions for static equilibrium?

The conditions for static equilibrium are that the net force acting on the object must be zero and that the net torque (rotational force) acting on the object must also be zero. These conditions ensure that the object will remain at rest and will not rotate.

3. How is static equilibrium different from dynamic equilibrium?

In static equilibrium, the object is at rest and has no acceleration, while in dynamic equilibrium, the object is in motion at a constant velocity. Additionally, in static equilibrium, the net force and net torque are both zero, while in dynamic equilibrium, there may be a non-zero net force or net torque acting on the object.

4. How is static equilibrium used in real-world applications?

Static equilibrium is used in many engineering and construction applications to ensure the stability and safety of structures. It is also used in physics experiments to measure forces and torques acting on objects.

5. What happens if an object is not in static equilibrium?

If an object is not in static equilibrium, it will either be in motion or will start to move due to the presence of a non-zero net force or net torque. This can lead to instability and potentially dangerous situations if not properly addressed.

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