Question

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

Answer 1

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

F = m a                  --> 1

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

m = V ρ                 --> 2

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

 

Hence the weight of the final casting is 4.84 kg