A cylindrical beaker of height 0.100{m} and negligible weight is filled to the b

Question

A cylindrical beaker of height 0.100{m} and negligible weight is filled to the brim with a fluid of density rho = 890{kg/m}^3 . When the beaker is placed on a scale, its weight is measured to be 1.00{N} A ball of density rho_b = 5000{kg/m}^3 and volume V = 60.0{cm}^3 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 = 9.81{m/s}^2 . What is the reading 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 F_r 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) Part D What weight W_3 does the scale now show?

Explanation / Answer

A cylindrical beaker of height 0.100m and negligible weight is filled to the brim with a fluid of density = 890kg/m^3 . When the beaker is placed on a scale, its weight is measured to be 1.00N . A ball of density = 5000kg/m^3 and volume = 60.0cm^3 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 = 9.81 m/s^2. What is the force 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). What weight does the scale now show? 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 A cylindrical beaker of height 0.100 {m} and negligible weight is filled to the brim with a fluid of density rho = 890 {kg/m}^3. When the beaker is placed on a scale, its weight is measured to be 1.00 {N}. A ball of density rho_b = 5000 {kg/m}^3 and volume V = 60.0 {cm}^3 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 = 9.81 {m/s}^2. What is the weight W_b of the ball? (The answer to this one is 2.94N) What is the reading W_2 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. (The answer to this one is 1.00N) What is the force F_r 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). (The answer to this one is 2.42 N) 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. What weight W_3 does the scale now show? he weight of the ball is rho*g*V = 5000 kg/m^3 * 9.8 m/s^2 * 0.00006 m^3 = 0.3 kg * 9.8 m/s^2 = 2.94 N Because the ball is being held up mostly by the rod, the fluid pressure on the bottom of the cylinder is just the same as before. The scale does not "know" the ball is there at all. That's why it still reads 1N. The buoyant force of the fluid on the ball is equal to the weight of the displaced fluid, namely, 890 kg/m^3 * 9.8 m/s^2 * 0.00006 m^3 = 0.52 N So the force needed for the rod to hold up the ball is 2.94 N - 0.52 N = 2.42 N Now the scale "feels" the weight of the ball, so the scale reads the weight of the ball PLUS the weight of the original fluid MINUS the weight of the fluid that was displaced = 2.94 N + 1.00 N - 0.52 N = 3.42 N

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