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Pressure, Buoyant Force problem 8

  • #1

Homework Statement


When a beaker partially filled with water is placed on an accurate scale the scale reads 22.0 g. If a piece of metal with a density of 3.800 kg/dm^3 and volume of 2.40 cm^3 is suspended by a thin string so that the metal is submerged in the water does not rest at the bottom of the beaker, what does the scale read

Homework Equations


Density=m/v
F=mg
Fb=density of fluid times volume of fluid times g
Fb=F(bottom) minus F(top)
f(apparent)=F(weight) minus Fb

The Attempt at a Solution


First I used the beaker that was partially filled with water and tried solving for the force
F=mg
F=.0022kg times 9.80 m/s^2
F=0.02156 N

Then I Solved for force again with the density and volume given but then I just realize how does that help me so idk what 2 do please help me?
 

Answers and Replies

  • #2
1,506
17
the submerged piece of metal experiences an UPTHRUST equal to the weight of water displaced (2.4cm^3 is displaced) (density of water = 1gram/cm^3)
Newton's 3rd law tells you that every action(force) has a reaction(force) which is equal and in the opposite direction.
This means that the UPTHRUST must produce a 'downthrust' (my term.... not a technical term)
Can you find what the balance will, read from this information?
 
  • #3
the submerged piece of metal experiences an UPTHRUST equal to the weight of water displaced (2.4cm^3 is displaced) (density of water = 1gram/cm^3)
Newton's 3rd law tells you that every action(force) has a reaction(force) which is equal and in the opposite direction.
This means that the UPTHRUST must produce a 'downthrust' (my term.... not a technical term)
Can you find what the balance will, read from this information?
If I multiply the density of water by the volume of the object (1000 kg/m^3 times 2.4 times 10^-6) then I get a mass of 0.0024 kg and Ik the answer has to be 24.4 g so Idk what Im doing wrong
 
  • #4
1,506
17
You have done it!!! 0.0024Kg is 2.4g. This is the Upthrust on the piece of metal and therefore the downthrust on the balance so the reading increases by 2.4g to give a reading of 24.4g
 
  • #5
Wow omg thanks :)
 

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