Chapter 15 44) As in the previous problem, a solid block is suspended from a spr

Question

Chapter 15

44) As in the previous problem, a solid block is suspended from a spring scale. If the reading on the scale when the block is completely immersed in water is 25.0 N, and the reading when it is completely immersed in alcohol of density 806 kg/m^3 is 25.7 N, what are (a) the block's volume and (b) its density?

Explanation / Answer

There are three forces acting on the block. gravity (downwards): Fg = m_block ·g = ? _block·V_block·g buoyancy (upwards) Fb = ? _fluid·V_block·g and the upward force of the suspension Fs All forces together are at equilibrium hence Fg + Fb + Fs = 0 => Fs = - Fg - Fb = (? _block - ? _fluid)·V_block·g Here we have measurement in two different fluids (i) Fsw = (? _block - ? _water)·V_block·g (ii) Fsa = (? _block - ? _alcohol)·V_block·g To find the volume of the block solve the equations for the unknown block density, equate them and then solve for the volume:: (i) => Fsw/(V_block·g) + ? _water = ? _block (ii) => Fsa/(V_block·g) + ? _alcohol = ? _block => Fsw/(V_block·g) + ? _water = Fsa/(V_block·g) + ? _alcohol ? _water - ?_alcohol = (Fsa - Fsw) / (V_block·g) => V_block = (Fsa - Fsw) / [ g · (? _water - ?_alcohol) ] = (Fsa - Fsw) / [ g · (? _water - ?_alcohol) ] = (25.7N - 25N ) / [ 9.81m/s² · (1000kg/m³ - 806kg/m³ ] = 3.678×10^-4m³ = 0.3678 L = 367.8 cm³ To get the density use volume and one of the two original equations: Fsw = (? _block - ? _water)·V_block·g ? _block = ? _water + Fsw / (V_block·g) = 1000kg/m³ + 25N / (3.678×10^-4m³ · 9.81m/s²) = 7928.6kg/m³

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