Buoyant suspension-let's skip Math

[SaintJude]

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Don't do the math (didn't give all constants you need)

Q: An anchor made of iron weights 833 N on the deck of a ship. If the anchor is now suspended in seawater by a massless chain, what is the tension in the chain?

A. 100 N
B. 724
C. 833 N
D. 957

1st: Does the keyword "suspended" indicate that acceleration of the anchor = 0?
If So, 1st equation would be T+Fb -W = ma= 0

Conceptually, does this mean that the tension is "carrying" only the weight not held up by the buoyant force? In a sense, the buoyant force is the "helping hand" to the tension.

Through that reasoning, I should have known that the tension would be slightly less than the anchor's weight, and thus answer choice B. Right reasoning?

 

[chiddler]

Sounds correct. Same thought process that I had.

 

[milski]

Yes. 👍

 

[MedPR]

Don't do the math (didn't give all constants you need)

Q: An anchor made of iron weights 833 N on the deck of a ship. If the anchor is now suspended in seawater by a massless chain, what is the tension in the chain?

A. 100 N
B. 724
C. 833 N
D. 957

1st: Does the keyword "suspended" indicate that acceleration of the anchor = 0?
If So, 1st equation would be T+Fb -W = ma= 0

Conceptually, does this mean that the tension is "carrying" only the weight not held up by the buoyant force? In a sense, the buoyant force is the "helping hand" to the tension.

Through that reasoning, I should have known that the tension would be slightly less than the anchor's weight, and thus answer choice B. Right reasoning?

Sounds good to me. The equation (for me) looks better written like this:

T+Fb=W. And to go a little further: The tension must be less than the weight, but not by much (e.g not by 733N) because in reality anchors are meant to sink in water, so the buoyant force must not be anywhere near enough to support the anchor by itself. Therefore, you would expect the tension to be greater than half the weight of the anchor.