A 4.0-kg cylinder of solid iron is supported by a string while submerged in water. What is the tension in the string? The density of iron is $7860 \mathrm {~kg} / \mathrm { m } ^ { 3 }$ and that of water is $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$
A) 34 N
B) 20 N
C) 40 N
D) 24 N
E) 2.5 N

Question

A 4.0-kg cylinder of solid iron is supported by a string while submerged in water. What is the tension in the string? The density of iron is $7860 \mathrm {~kg} / \mathrm { m } ^ { 3 }$ and that of water is $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$
A)  34 N
B)  20 N
C)  40 N
D)  24 N
E)  2.5 N

Quizplus · Accepted Answer

The buoyant force on the cylinder is equal to the weight of the water displaced by the cylinder. The weight of the water displaced is equal to the volume of the cylinder times the density of water times the acceleration due to gravity. The volume of the cylinder is equal to the mass of the cylinder divided by the density of iron. Therefore, the buoyant force is equal to:$$F_b = \rho_{water} V g = \rho_{water} \frac{m_{iron}}{\rho_{iron}} g$$The tension in the string is equal to the weight of the cylinder minus the buoyant force. Therefore, the tension in the string is:$$T = m_{iron} g - F_b = m_{iron} g - \rho_{water} \frac{m_{iron}}{\rho_{iron}} g$$$$T = m_{iron} g \left(1 - \frac{\rho_{water}}{\rho_{iron}}\right)$$$$T = (4.0 \mathrm{~kg}) (9.8 \mathrm{~m/s^2}) \left(1 - \frac{1000 \mathrm{~kg/m^3}}{7860 \mathrm{~kg/m^3}}\right)$$$$T = 34 \mathrm{~N}$$

A 40-Kg Cylinder of Solid Iron Is Supported by a String