The rod is now shortened and attached to the bottom of thebeaker. The beaker is

Question

The rod is now shortened and attached to the bottom of thebeaker. The beaker is again filled with fluid, the ball issubmerged and attached to the rod, and the beaker with fluid andsubmerged ball is placed on the scale. (Intro2 figure)

please help with c and d; numerical values would beappreciated.

(ps: ans for a=2.94 N and b=1 N)

thank you

A cylindrical beaker of height 0.100 m and negligible weight is filled to thebrim with a fluid of density Rho = 890 kg/m^3 . When the beaker isplaced on a scale, its weight is measured to be 1.00 N . (Intro1 figure) A ball of density Rhob = 5000 Kg/m^3 and volume V = 60.0 cm^3 is then submerged inthe fluid, so that some of the fluid spills over the side of thebeaker. The ball is held in place by a stiff rod of negligiblevolume and weight. Throughout the problem, assume the accelerationdue to gravity is g = 9.81 m/s^2 . A) What is the weight Wb of the ball? (in N) B) What is the reading W2 of the scale when the ball is held inthis submerged position? Assume that none of the water that spillsover stays on the scale. ? C) What is the force Fr applied to the ball by the rod? Takeupward forces to be positive (e.g., if the force on the ball isdownward, your answer should be negative).? The rod is now shortened and attached to the bottom of thebeaker. The beaker is again filled with fluid, the ball issubmerged and attached to the rod, and the beaker with fluid andsubmerged ball is placed on the scale. (Intro2 figure) D) What weight W3 does the scale now show? please help with c and d; numerical values would beappreciated. (ps: ans for a=2.94 N and b=1 N)

Explanation / Answer

c) Here the weight acts downwards and the tension and the Buoyantforce acts in the upward direction Therefore at the equilibrium we know that the force exerted bythe ball on the rod is given by                      FB +T =mg                              T =mg -FB                                  = mg -gV                                  = 2.94N -(890kg/m3)(9.81m/s2)(60*10-6m3)                                  = 2.94N -0.523854N                                   =2.42N d) When the rod is attached to the bottom then tension and weightacts downwards and buoyant force acts upwards then the equation isgiven by                       T +mg =FB                                T=FB -mg                                  = 0.523854N-2.94N                                  = -2.42N The weight W3 on thescale is nothing but the tenison which is equal in magnitudewith the first case acting upwards i.e T = 2.42N + 1.00N =3.42N ( part c + beaker weight should beadded ) The negative sign indicates that the force is applied on thebottom.                                T=FB -mg                                  = 0.523854N-2.94N                                  = -2.42N The weight W3 on thescale is nothing but the tenison which is equal in magnitudewith the first case acting upwards i.e T = 2.42N + 1.00N =3.42N ( part c + beaker weight should beadded ) The negative sign indicates that the force is applied on thebottom.

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