Question

A wooden ball with a weight of 18.0 N hangs from a string tied to a spring scale. When the ball is at rest, exactly 50% submerged in water, the spring scale reads 5.00 N. For this problem, we will use a density of water of 1000 kg/m3, and we will use g = 10.0 m/s2. (a) If the ball was only 20.0% submerged in water instead, what would the spring scale read? (b) Determine the density of the ball. (c) Determine the volume of the ball.

Answer 1

Answer:

a) $12.8N$

b) $692.308\frac{kg}{m^{3}}$

c) $2.6(10^{-3})m^{3}$

Explanation:

Let's write the information given by the exercise.

$W(WoodenBall)=18.0N$

δ (Water) = $1000\frac{kg}{m^{3}}$

$g=10.0\frac{m}{s^{2}}$

The buoyant force exerted by the water is :

E= V.δ.g

Where E is the buoyant force

V is the shifted volume of the fluid

δ is the fluid density

and g is the gravity acceleration.

We can write the following equation :

$W(WoodenBall)-E=5.00N$

18.0 N - V.δ(water).g = 5.00 N

$13N=0.5[V(WoodenBall)]1000\frac{kg}{m^{3}}.10\frac{m}{s^{2}}$

Let's remember that

$1N=1\frac{kg.m}{s^{2}}$

Therefore,

$13N=0.5[V(WoodenBall)].10000\frac{N}{m^{3}}$

$V(WoodenBall)=2.6(10^{-3})m^{3}$

And that is the answer for c)

For a) we write

$SpringScaleReading=W(WoodenBall)-E$

Spring scale reading = 18 N - 0.2 [V(Wooden ball)] . δ(water). g

$SpringScaleReading=18N-0.2[2.6(10^{-3})m^{3}].1000\frac{kg}{m^{3}}.10\frac{m}{s^{2}}$

$SpringScaleReading=18N-5.2N=12.8N$

For a) The spring scale will read 12.8 N

b) $W(WoodenBall)=m(WoodenBall).g$

Where m(Wooden Ball) is the mass of the wooden ball.

$18N=m(WoodenBall).10\frac{m}{s^{2}}$

$m(WoodenBall)=1.8kg$

δ(Wooden ball) = m(Wooden ball) / V(Wooden ball)

δ(Wooden ball) = 1.8 kg / $2.6(10^{-3})m^{3}$

δ(Wooden ball) = $692.308 \frac{kg}{m^{3}}$