A cylindrical beaker of height 0.100m and negligible weight isfilled to the brim

Question

A cylindrical beaker of height 0.100m and negligible weight isfilled to the brim with a fluid of density = 890kg/m3 .When the beaker is placed on a scale, its weight is measured to be1.00N .
A ball of density = 5000kg/m3 and volume =60.0cm3 is then submerged in the fluid, so that some ofthe fluid spills over the side of the beaker. The ball is held inplace by a stiff rod of negligible volume and weight. Throughoutthe problem, assume the acceleration due to gravity is g = 9.81m/s2.

What is the force applied to the ball by the rod? Take upwardforces to be positive (e.g., if the force on the ball is downward,your answer should be negative).
The rod is now shortened andattached to the bottom of the beaker. The beaker is again filledwith fluid, the ball is submerged and attached to the rod, and thebeaker with fluid and submerged ball is placed on thescale.
What weight does the scale now show?
The rod is now shortened andattached to the bottom of the beaker. The beaker is again filledwith fluid, the ball is submerged and attached to the rod, and thebeaker with fluid and submerged ball is placed on thescale.
What weight does the scale now show?

Explanation / Answer

a) 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 b) 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. 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.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.