Static Equilibrium: Why Ethelred Gets the Short End of the Pole

[PhysicsKid0123]

It is a thousand yeas ago. You, along with your friend king Ethelred the Unready, have been hunting wild boar. You've had good luck; you shot a 150 lb boar. Now the hard part; you have to carry it back to the castle several furlongs away. Fortunately you have some rope and an 8 cubit pole. You will suspend the boar from the pole and the two of you will support the pole on your shoulders. Ethelred will support the front of the pole; you will support the back. You convince Ethelred that the boar should not be suspended right in the middle of the pole, but for better steering of the pole the boar should be suspended closer to Ethelred than to you. In a single well written paragraph, using physics, explain why Ethelred is getting a bad deal.

 

[SteamKing]

Well, we aren't going to write the paragraph for you, or do the physical calculations.

 

[PhysicsKid0123]

I need someone to check my answer: "W = F *d where d is the distance to the center of mass of the system. since the center is closer to him, he will have to apply a greater force in order to do the same work as yours. This is just like using a wedge to lift something. If you use the wedge, you will need a much less force to lift it than you would need if you are doing it by lifting it with no wedge."

This right?

 

[PhanthomJay]

You can't use the work analogy because no work is being done . Try using the equilibrium equations to determine who applies the greater force , and write the sentence accordingly.

 

[haruspex]

You can't use the work analogy because no work is being done . Try using the equilibrium equations to determine who applies the greater force , and write the sentence accordingly.

I'm sure you're right in terms of what's wanted as the answer, but it's interesting to ask why it matters if no work is being done. In practice, work is done. You cannot even hold up a heavy object, let alone walk carrying it, without doing work, and the greater the force needed, the more work. The effective distance, in the W=Fd sense, probably does not depend on F.