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3 years ago

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A sand core used to form the internal surfaces of a steel casting experiences a buoyancy force of 225 n. the volume of the mold

cavity forming the outside surfaces of the casting = 4840 cm3. what is the weight (in kg) of the final casting? ignore considerations of shrinkage

1 answer:

pashok25 [27]3 years ago

5 0

Newton’s second law gives us the relationship of force F, mass m and acceleration a. The formula is given as:

<span>F = m a --> 1</span>

However we also know that the relationship of mass m, density ρ, and volume V is:

<span>m = V ρ --> 2</span>

Therefore substituting equation 2 to equation 1:

F = ρ V a = ρ V g

where a is acceleration due to gravity, ρ is density of water and V is the volume of the casting, therefore:

F = (1x10^-3 kg/cm^3) (4840 cm^3) (9.8 m/s^2)

F = 47.432 kg m/s^2

F = 47.432 N

Going back to equation 1:

47.432 N = m (9.8 m/s^2)

m = 4.84 kg

<span>Hence the weight of the final casting is 4.84 kg</span>

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