buoyant force/apparent weight

[greenseeking]

Hello

I was wondering if someone could clear up a concept for me.

Question: An object weight 96 N is suspended by string while completely immersed in water. the object has a density three times the density of water. What is its apparent weight?

Answer: 64N

I got the right answer by going: Mg-Fb=96N-96/3=64N. However, what happens to the Tension of the string? Shoudn't it be Mg-Fb-T?

There are two "up" forces, tension and Fb (Bouyant Force) and one "down" force, weight. When I'm trying to find the apparent weight, shouldn't I subtract the force of Tension as well? I guess my question is why didn't the answer not take tension into account? the string is what is helping the object stay afloat, since the density is three times greater than water right? Wouldn't the object want to sink right down to the floor without the string?

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[milski]

The tension force has the same magnitude as the apparent weight. That's precisely why you got the right answer. T+Fb=Mg or T=Mg-Fb.

T in the equation above is the tension force exerted on the block by the string. That and the apparent weight, exerted by the block on the string, are a 3rd law pair - same magnitude, opposite directions and swapped on/by bodies.

 

[greenseeking]

thank you milski! Makes sense now... 3rd law pair...i gotta remember the basics!

 

[cheff949]

How come you divided by 3?

 

[milski]

How come you divided by 3?

The object has 3 times the density of water. Water with the same volume as the object will weight 1/3 of the weight of the object.

 

[Tatiana3325]

Hello

I was wondering if someone could clear up a concept for me.

Question: An object weight 96 N is suspended by string while completely immersed in water. the object has a density three times the density of water. What is its apparent weight?

Answer: 64N

I got the right answer by going: Mg-Fb=96N-96/3=64N. However, what happens to the Tension of the string? Shoudn't it be Mg-Fb-T?

There are two "up" forces, tension and Fb (Bouyant Force) and one "down" force, weight. When I'm trying to find the apparent weight, shouldn't I subtract the force of Tension as well? I guess my question is why didn't the answer not take tension into account? the string is what is helping the object stay afloat, since the density is three times greater than water right? Wouldn't the object want to sink right down to the floor without the string?

The object is not floating. It is submerged in water, is not accelerating and must be suspended by a string (because the object is more dense than the water) with tension equal to the apparent weight of the object. If the same object were in air instead of water, the tension would be equal to the objects weight.