Solution - American Association of Physics Teachers

2013 F = ma Exam 1315. A uniform rod is partially in water with one end suspended, as shown in figure. The density of the rod is 5/9 thatof water. At equilibrium, what portion of the rod is above water?0000000000000000011111111111111111000000000000000001111111111111111100000000000000000111111111111111110000000000000000011111111111111111000000000000000001111111111111111100000000000000000111111111111111110000000000000000011111111111111111000000000000000001111111111111111100000000000000000111111111111111110000000000000000011111111111111111(A) 0.25(B) 0.33(C) 0.5(D) 0.67(E) 0.75SolutionSuppose a fraction α of the rod is above water. Let the length of the rod be l, the volume of the rod be V ,the density of water be ρ w , and the density of the rod be ρ r . Consider torques about the pivot point. Gravityapplies a torqueτ g = ρ r V g · 12 lA fraction 1−α of the rod is submerged, and the center of the submerged portion is a distance (α+ 1 2 (1−α))lfrom the pivot. So the buoyant force applies a torqueτ b = ρ w (1 − α)V g ·(α + 1 )2 (1 − α) lThese torques must balance:τ b = 1 2 ρ w(1 − α 2 )V glτ g = τ b12 ρ r = 1 2 ρ w(1 − α 2 )ρ rρ w= 1 − α 2We are given ρrρ w= 5 9 , so that 59 = 1 − α2α = 2 3Copyright c○2013 American Association of Physics Teachers