Consider the following diagram below. The center of mass of the meter stick is at the center (50 cm from the left hand end). It is supported by a string at the stick. When a load of mass is 50 grams is hanging from the meter stick at the 85cm position on the stick, the meter stick is in equilibrium. The load is then completely suspended in a container of fluid whose density is rho fluid = 800 g/ml. This causes the support to be relocated to the 70 cm mark stick to achieve static equilibrium again. What must be the density of the block?
Explanation / Answer
Give data The center of mass of the rod is xcm = 50 cm Support by a string of the rod x1 = 75 cm the mass of the load mblock = 50 g The load hang from the stick x2 = 85 cm the density of the fluid fluid = 800 g/ml ---------------------------------------------------------------------------------------------------------------- Consider pivot at left end then by appliying principle of moments let the mass of the stick mstick mstick (50 cm) - T(75 cm) + mblock(85 cm) = 0 . . . . . . (1) Since it is static equilibrium, the net force is zero mstick g - T + mblockg = 0 T = mstick g + mblockg from above equation (1) mstick (50 cm) - [mstick g + mblockg ](75 cm) + mblock(85 cm) =0 mstick g (-25 cm) + mblockg (10cm) = 0 mstick = mblock (10cm)/ (25 cm) mstick = (50 g) (10cm)/ (25 cm) = 20 g --------------------------------------------------------------------------------------------------------- When block is immersed in fluid then the block is hang at 70 cm mark so mstick (50 cm) - T(75 cm) + F(70 cm) = 0 (20 g) (50 cm) - (20 g + 50g) (980 cm/s2) (75 cm) + F(70 cm) = 0 F(70 cm) = 68600 - 1000 = 67600 F = 67600/70 = 965.7 dynes Here F is the apparent weight of the block so F = mg - FB 965.7 = (50 g)(980 ) - Vg 965.7 = (50 g)(980 ) - V(800)(980) V(800)(980) = 49000 - 965.7 V = 0.061 cm3 Therefore the density of the block is = m/V = 50 g / 0.061 = 819 mL or 819 cm3 Therefore the density of the block is = m/V = 50 g / 0.061 = 819 mL or 819 cm3