SOLUTION (A) Find the normal force on the cube when half-immersed in water. MA 1
ID: 1511751 • Letter: S
Question
SOLUTION (A) Find the normal force on the cube when half-immersed in water. MA 1.00x10-kg-= 0.370 m Calculate the volume V of the cube and the length d of one side, for future reference (both quantities will be needed for what follows) Val = 1.00×103 kg PA 2.70 x 103 kg/m5 d = Val1/3 = 0.718 m n-MAIgBwater n = MAS-Bwater-Mag-pwater(V/2)g Write Newton's second law for the cube, and solve for the normal force The buoyant force is equal to the weight of the displaced water (half the volume of the cube) = ( 1.00 x 103 kg/m3 )(9.80 m/s2 -( 1.00 × 103 kg/m3 )(0.370 m3/2.00 )( 9.80 m/s2 n = 9.80× 103 N-1.81 X 103 N = 7.99 x 103Explanation / Answer
In euilibrium
Fnet = 0
n + FBw + FBG - Mg = 0
n = Mg - FBw - FBG
density of aluminium = Dal = 2700 kg/m^3
M = mass of aluminium M = V*Dal
FBw = Dw*V/3*g
FBG = DG*2V/3*g
n = Dal*V*g - Dw*V/3*g - DG*2V/3*g
n = ((2700 - (1000/3) - (1260*2/3))*1.5^3/9.8
n = 525.7 N <<<<----answer
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