Question

Three objects of identical mass attached to strings are suspended in a large tank of liquid, as shown.

(a) Must all three strings have the same tension? 

____ Yes ____ No 

Justify your answer. 

Object A has a volume of 1.0 x 10^-5 m³ and a density of 1300kg m³  The tension in the string to which object A is attached is 0.0098 N.

(b) Calculate the buoyant force on object A. 

(c) Calculate the density of the liquid. 

(d) Some of the liquid is now drained from the tank until only half of the volume of object A is submerged. 

Would the tension in the string to which object A is attached increase, decrease, or remain the same? 

____ Increase ____ Decrease ____ Remain the same

Answer 1

(a) There are three forces acting on the masses in each case. Tension up, buoyant force up, weight down. Since they are at rest we have. 

F_net = 0 

F_t + F_b = mg 

F_t = mg – F_b 

so the largest F_b makes the largest F_t

We are to assume the diagram is to scale and that clearly the volumes of the three containers are different. The one with the largest volume displaces the largest amount and weight of water and will have the largest buoyant force acting on it. So since they all displace different volumes (and weights) of water they all have different buoyant forces, and based on the equation shown above will have different tensions.

(b) The mass of the object is given by m = ρ_objV_obj. 

Using the equation from part a, 

 F_t + F_b = mg, F_t + F_b = (ρ_objV_obj.)g (0.0098) + F_b = (1300)(1 x 10^–5)(9.8) 

F_b = 0.1176N 

(c) The buoyant force is by definition equal to the weight of the displaced fluid.  

F_b = (ρ_fluidV_disp.) g 

0.1176 = ρ_fluid (1 x 10^–5)(9.8) ρ_fluid = 1200kg/m³ 

(d) With only half of the volume submerged, ½ as much water will be displaced and the buoyant force will be half the size. Based on the formula from part A, less buoyant from will make a larger tension. This also makes sense conceptually. Objects have large apparent weights in air than water so having some of it in the air will increase its apparent weight.