# Archimedes' Principle problem help please

1. 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 2. Homework Equations
Fb=pVg
F=ma

3. 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?