Question

In: Physics

A cylindrical beaker of height 0.100 m and negligible weight is filled to the brim with a fluid of density ρ = 890 kg/m3 . When the beaker is placed on a scale, its weight is measured to be 1.00 N

A cylindrical beaker of height 0.100 m and negligible weight is filled to the brim with a fluid of density ρ = 890 kg/m3 . When the beaker is placed on a scale, its weight is measured to be 1.00 N .(Figure 1)

A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 .

Part A

What is the weight Wb of the ball?
Express your answer numerically in newtons.

Part B

What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale.
Calculate your answer from the quantities given in the problem and express it numerically in newtons.

Part C

What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative).
Express your answer numerically in newtons.

Part D

What weight W3 does the scale now show?
Express your answer numerically in newtons.

 

Solutions

Expert Solution

A)

The weight of the ball is rho*g*V
= 5000 kg/m^3 * 9.8 m/s^2 * 0.00006 m^3
= 0.3 kg * 9.8 m/s^2 = 2.94 N

B)

Because the ball is being held up mostly by the rod,
the fluid pressure on the bottom of the cylinder is just the same as before.
The scale does not "know" the ball is there at all.
That's why it still reads 1N.

C)

The buoyant force of the fluid on the ball is equal to the weight of
the displaced fluid, namely,
890 kg/m^3 * 9.8 m/s^2 * 0.00006 m^3
= 0.52 N
The force needed for the rod to hold up the ball is 2.94 N - 0.52 N
= 2.42 N.

D)

Now the scale "feels" the weight of the ball,
so the scale reads the weight of the ball
PLUS the weight of the original fluid
MINUS the weight of the fluid that was displaced
= 2.94 N + 1.00 N - 0.52 N
= 3.42 N


A)

= 2.94 N

B)

It still reads 1N.

C)

= 2.42 N

D)

= 3.42 N

Related Solutions