A metal block of mass 160 g rests in equilibrium at the bottom of a glass of water, contacting the base at a few points. Given the metal's density is 8000 kg/m³, determine the normal force the base exerts on the block.

1.4 N
2.7 N
3.3 N
3.8 N

Correct answer: 1.4 N

Solution

Volume V = m/rho_metal = 0.160/8000 = 2e-5 m³. Weight = m*g = 0.160*9.8 = 1.568 N. Buoyancy = rho_water*V*g = 1000*2e-5*9.8 = 0.196 N. Normal force N = weight - buoyancy = 1.568 - 0.196 = 1.372 N ~ 1.4 N.

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