A cylindrical beaker of height 0.100m and negligible weight is filled to the bri
ID: 1312960 • Letter: A
Question
A cylindrical beaker of height 0.100m and negligible weight is filled to the brim with a fluid of density = 890kg/m3 . When the beaker is placed on a scale, its weight is measured to be 1.00N .(Figure 1) A ball of density b = 5000kg/m3 and volume V = 60.0cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81m/s2 .
What is the weight Wb of the ball?
Express your answer numerically in newtons.
What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale.
Calculate your answer from the quantities given in the problem and express it numerically in newtons.
What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative).
Express your answer numerically in newtons.
The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.(Figure 2)
What weight W3 does the scale now show?
Express your answer numerically in newtons.
Explanation / Answer
This is a typical Archimedes Principle problem:
A body submerged in a fluid (liquid) is pushed up with a force equal to the weight of the liquid displaced.
The problem wants you to find the buoyant force (in Newtons)
So, your displaced liquid is 60mL bcz 1 cm^3=1mL
Liquid density is 5 g/cm^3 . Considering that Weight = m*g,
This gives me an answer of 0.524 N exercised upwards on the rod.
The weight of your body is 0.3 Kg, so W = 0.3 * 9.81 = 2.42 N
The scale should show = 2.42N + 1N - 0.524 N = 2.89 N
Good luck. Doublecheck, please
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