Question

What does the lower scale read? Answer in units of N

Answer 1

We will have the following:

First, we are given:

*Mass of the breaker: 1.1kg

*Mass of water: 3.3 kg

*Mass of metallic alloy: 4.2kg

*Density of the alloy: 5300kg/m^3

*Density of water: 1000kg/m^3

Now, we find the volume of water displaced by the alloy:

$V_{\text{w}}=4.2\operatorname{kg}\cdot\frac{m^3}{5300\operatorname{kg}}\Rightarrow V_w=\frac{21}{26500}m^3\Rightarrow V_w\approx7.92\cdot10^{-4}m^3$

Then, from the reading in the hanging scale we will have the force experienced by the alloy due to the upthrust when placed in water, that is:

$R=mg-\rho Vg$

So:

$R=(4.2\operatorname{kg})(9.8m/s^2)-(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)$$\Rightarrow R=33.39396226\ldots N\Rightarrow R\approx33.4N$

The reading on the lower scale is due to the weight of the water in the breaker and upthrust on the scale:

$R=g(m_1+m_2)+\rho Vg$

Finally:

$R=(9.8m/s^2)(1.1\operatorname{kg}+3.3\operatorname{kg})+(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)$$\Rightarrow R=50.886003774\ldots N\Rightarrow R\approx50.9N$

So, the readin on the lower scale is approximately 50.9N.