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Problem Sets:
Problem
There is a solid spherical object. When weighed in air using a spring scale, the scale reads 12 N. When half of the object’s volume is immersed in water, the spring scale reads 5 N. If the object is then gently released into the water so that it floats freely at equilibrium, the buoyant force acting on the object will be:
Hints
Hint 1:
The buoyant force is given by the principle of flotation, $F_{B, \text{float}} = W$.
$$F_{B, \text{float}} = 12 \text{ N}$$The correct option is D.
Let $W$ be the true weight of the solid spherical object. Let $F_{B, \text{float}}$ be the buoyant force acting on the object when it floats freely in equilibrium.
[Q1]
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Determine the object's weight: When the object is weighed in air, the spring scale reading directly measures its weight, assuming the buoyancy of air is negligible.
$$W = 12 \text{ N}$$ -
Apply the principle of flotation: When an object is released and floats freely at equilibrium, it is stationary. According to Newton's first law, the net force on the object is zero. The forces acting on the floating object are its weight ($W$) acting downwards and the buoyant force ($F_{B, \text{float}}$) acting upwards.
The condition for static equilibrium is:
$$\sum F_y = F_{B, \text{float}} - W = 0$$ -
Calculate the buoyant force: From the equilibrium condition, the buoyant force must be equal in magnitude to the object's weight.
$$F_{B, \text{float}} = W$$Substituting the known value of the weight:
$$F_{B, \text{float}} = 12 \text{ N}$$The information that the scale reads 5 N when half the object is immersed is consistent with this result. It shows that the maximum buoyant force (when fully submerged) would be $2 \times (12 \text{ N} - 5 \text{ N}) = 14$ N. Since the weight (12 N) is less than the maximum buoyant force (14 N), the object will indeed float.