How Does Archimedes' Principle Apply to Forces on a Submerged Cube?

AI Thread Summary
The discussion focuses on applying Archimedes' Principle to a submerged cube in a liquid. Participants are tasked with calculating the forces acting on the cube, including the total downward force on the top, the upward force on the bottom, and the tension in the rope. The correct values for the downward force and buoyant force are provided, while the upward force calculation remains unclear for some users. A hint emphasizes that the forces on the top and bottom are related, and participants seek clarification on specific variables and equations involved in the calculations. Understanding these relationships is crucial for solving the problem accurately.
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Homework Statement


In the figure below, a cube of edge length L = 0.500 m and mass 480 kg is suspended by a rope in an open tank of liquid of density 1030 kg/m3.
(a) Find Ftop, the magnitude of the total downward force on the top of the cube from the liquid and the atmosphere, assuming atmospheric pressure is 1.00 atm.
correct check mark 2.59E4N

(b) Find Fbottom, the magnitude of the total upward force on the bottom of the cube.
wrong check mark N

(c) Find T, the tension in the rope.
N

(d) Calculate Fbu, the magnitude of the buoyant force on the cube using Archimedes' principle.
correct check mark 1.26E3N
What relation exists among all these quantities? (Select all that apply.)
Fbu = Fbottom + T
Fbu = m - T
Fbu = Ftop - Fbottom
Fbu = Fbottom - Ftop
Fbu = T - m

wrong check mark

hint:Force is equal to the product of the (uniform) pressure and the face area. The gauge pressure at a certain depth h in a fluid is equal to ρgh. The gravitational force and the force from the rope also act on the cube. By Newton's second law, the net force must be zero.
Section 14-7 Archimedes' Principle
hrw7_14-37.gif

Homework Equations


Fb=pVg
F=ma

The Attempt at a Solution


I have no idea what to do for part b
at first I thought Fbottom was the same as the buoyancy force but that was wrong.
All I need is a hint on how to start part b and then I should be able to get part c on my own. I have already solved for parts A and D
 
Last edited:
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Think about what caused the force on the top in part A. It is the same type of force on the bottom.
 
A(Po+pg(3/2)L)=2.71E4N
thanks
 
would T=Ftop+mg-Fb-Fbottom ?
 
Can someone help me solve A and B for this problem..?

A(Po+pg(3/2)L)=2.71E4N

what does Po and pg mean in here?
is A = 1.5 in this problem?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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