1) Force read on the upper scale: 33.4 N
2) Force read on the lower scale: 43.0 N
Explanation:
1)
The reading on the upper scale is equal to the net force acting on the block of metallic alloy. The net force is given by:
[tex]F=W-B[/tex] (1)
where
W is the weight of the block (downward)
B is the buoyant force (upward)
The weight of the block is given by
[tex]W=mg[/tex]
where m = 3.7 kg is the mass of the block and [tex]g=9.8 m/s^2[/tex] is the acceleration of gravity.
The buoyant force is given by
[tex]B=\rho_w V g[/tex]
where
[tex]\rho_w=1000 kg/m^3[/tex] is the water density
V is the volume of the block
The volume of the block can be written as
[tex]V=\frac{m}{\rho_b}[/tex]
where [tex]\rho_b=4600 kg/m^3[/tex] is the density of the block.
Substituting everything into eq.(1), we find:
[tex]F=mg-\rho_w \frac{m}{\rho_b}g=(3.7)(9.8)-(1000)\frac{1.3}{4600}(9.8)=33.4 N[/tex]
2)
Here we want to find the force on the lower scale.
The force on the lower scale is equal to the difference between the total weight of the system (given by the weight of the beaker + the weight of the water + the weight of the block) and the upper net force exerted on the upper scale, therefore:
[tex]F' = m_B g + m_wg+m_b g - F[/tex]
where:
[tex]m_B=1.3 kg[/tex] is the mass of the beaker
[tex]m_w=2.8 kg[/tex] is the mass of the water
[tex]m_b = 3.7 kg[/tex] is the mass of the block
[tex]F=33.4 N[/tex] is the upper net force
Substituting and solving, we find:
[tex]F'=(1.3+2.8+3.7)(9.8)-33.4=43.0 N[/tex]
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