Normal Force in Water and Finding Density

[CaityAnn]

Hello,
Im having trouble with this chapter, So thanks for your patience.

Heres how the first question is stated:
1) A 30.0 kg irregularly shaped rock with a density of 4600 kg/m^3 rests on the bottom of a swimming pool.
What is the normal force of the pool bottom on the rock?

What have I tried?
well the normal force should be equal to the force exerted back onto the rock from the bottom of the pool. Also there should not be any air under the pool so the atm shouldn't mattter.

2) You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 N. If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17.0 N.
What is the density?

What have I tried?
Density= m/V. m= 28.4/9.8=2.898 kg.
To find the volume use 17N/9.8=1.735kg according to the water but i think this may be wrong. Anyways, if this was right, take 1.735/v=1000, now i have a V and a M, solve for density of statue. This is wrong I believe so please tell me what's wrong with my thought process.

 

[arunma]

Hello Ann. I think I can help you here.

Heres how the first question is stated:
1) A 30.0 kg irregularly shaped rock with a density of 4600 kg/m^3 rests on the bottom of a swimming pool.
What is the normal force of the pool bottom on the rock?

What have I tried?
well the normal force should be equal to the force exerted back onto the rock from the bottom of the pool. Also there should not be any air under the pool so the atm shouldn't mattter.

You're right. The normal force is equal to the force exerted back onto the rock from the bottom of the pool; in fact this is the definition of normal force. So we must ask: what is the force that the pool exerts on the rock? By Newton's third law, the fact that the bottom of the pool is level means that the normal force is equal to the force that the rock exerts on the bottom of the pool (remember, the rock and the pool exert forces on each other).

You're probably familiar with buoyant force, so you know that any immersed object has a buoyant force acting on it. What are the forces acting on the rock? Think about how you would solve this problem if there were no water, and then think about the effect of buoyant force.

2) You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 N. If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17.0 N.
What is the density?

What have I tried?
Density= m/V. m= 28.4/9.8=2.898 kg.
To find the volume use 17N/9.8=1.735kg according to the water but i think this may be wrong. Anyways, if this was right, take 1.735/v=1000, now i have a V and a M, solve for density of statue. This is wrong I believe so please tell me what's wrong with my thought process.

Here you need to use Archimedes' principle. You correctly found the mass of the statue, so you only need its density. From the spring scale readings given in the problem, you should be able to determine the buoyant force from the information in the problem. Now can you use Archimedes' principle to find the volume of the statue?

Well, I hope that helps. Feel free to ask if you're still having trouble. Also, beware of a common mistake that students often make when using Archimedes' principle. Remember that the density refers to the density of the fluid, not the density of the object.

 

[CaityAnn]

I do still need help.
According to Archimedes the Bouyancy force equals the weight of the fluid displaced by the object. So w or Fb is equal to P(fluid)*Volume(fluid)*g... So I need a volume for fluid found using M(object)=P(object)*V(object).
So... 30kg/4600=.0065. Then .0065*1000*9.8=63.7, and according to the odd problem in back of book that is way wrong. I am thinking maybe I should be adding on a missing factor... unsure.

 

[CaityAnn]

Part 2.
To find the volume I did 17N=1000*V*9.8, solved for V and got .002.
Took 2.898/.002=1450 And this didnt work. <discouraged>

 

[Gokul43201]

Hello,
Im having trouble with this chapter, So thanks for your patience.

Heres how the first question is stated:
1) A 30.0 kg irregularly shaped rock with a density of 4600 kg/m^3 rests on the bottom of a swimming pool.
What is the normal force of the pool bottom on the rock?

What have I tried?
well the normal force should be equal to the force exerted back onto the rock from the bottom of the pool.

Actually, the normal force IS the force exerted upwards by the pool floor on the bottom surface of the rock.

I do still need help.
According to Archimedes the Bouyancy force equals the weight of the fluid displaced by the object. So w or Fb is equal to P(fluid)*Volume(fluid)*g... So I need a volume for fluid found using M(object)=P(object)*V(object).
So... 30kg/4600=.0065. Then .0065*1000*9.8=63.7, and according to the odd problem in back of book that is way wrong. I am thinking maybe I should be adding on a missing factor... unsure.

You're on the right track so far - good job.

First thing you need to do in any force problem is draw the Free Body Diagram. What are the forces acting on the rock?

There's gravity (it's weight) acting downwards; there's the buoyant force (Fb, that you've just calculated) acting upwards, and there's a normal reaction acting upwards. Using the fact that the rock is not accelerating in the vertical direction, can you now find the size of the normal reaction (using Newton's second law)?

PS: Once you've done this and got the hang of it, the second problem is simply an application of the same concept (and working backwards). Replace the pool floor by the scales (instead of a spring scale, imagine your bathroom scales lying at the bottom of the tub of water) and the normal force becomes the reading on the scales.

 

[CaityAnn]

Thanks for your reply.
1) I tried to get away with not making a FBD.. shouldn't have.
2) I am trying to figure out the second still. So I know the mass. If the statue is on a scale on the bottom of the tub and the scale reads 17N, this would be the Fb and like the previous problem n=-Fb+w. But I can't figure out the point of that.
Im thinking the displacement of water should be related to the density of the object. From 28.4N-17N the difference of 11.4 N should mean something.
Doing doing my FBD.. the n is equal to 11.4 as well... Ok so the normal force is the pressure ? felt by the block now.. I need help now. :)

 

[CaityAnn]

So 11.4 is the N force of the pool pushing back up. and I know Pressure(water)=F(normal)/A(volume)?