Question

If we combine the downward force of gravity with the upward buoyant force, we get what is known as the apparent weight of the object, Wapparent = mobject9 – Fbuoyant = PobjectgVtotal – Poutside gVdisplaced · For our experiment this week, we will be looking at a mass submerged in water. The apparent weight will be the weight force due to gravity minus the buoyant force due to the surrounding water. Consider the setup to the right with an object mobject = 0.500 kg submerged in water (water = 1000 kg/m' = 1.00 gram/cm) and suspended at rest from a string. The object is cylindrical with radius r= 2.00 cm and height h = 6.00 cm. a) [2 points] Draw a free-body diagram for the mass. What is the net force on the mass? b) [3 points] What is the volume of the mass? Determine the buoyant force on the mass. c) [2 points] Determine the apparent weight of the mass and the tension force in the string. d) [2 points] Keeping the volume V of the mass constant, at what mass m would the tension force be reduced to zero? [Above this mass, the object would sink without the string and below the mass, the object would float.]

Answer 1

# Given mobiect 50 Mobject -0.500kg Pw = 1000lg/m3 = 1 g/cm3 8=2cm, h=6cm - FB HT FiBiD: um Y W Het force=w - FB Volume = *(2) 2x 6 = 75.3982 cm3. FB = (Volume) Pw g = (75.3982x10-5) x1000x9.81 +0.7396N O App. wt = mg - FB - (0.500X9.81-0.7396) N = 4.1654N @ Net Force zo ng=FB - 0.7396 0.0754 kg 9.81