91Ó°ÊÓ

In this case, the scales have negligible weight. Thus, the total force to be supported by the four scales is the weight of the sack (\(220.0\mathrm{N}\)). Since the scales are identical and all four scales show the same reading, the force on each scale is equal. Therefore, the force on each scale (F) is: \(F=\frac{220.0\:\mathrm{N}}{4}=55.0\:\mathrm{N}\).

In this case, each scale has a weight of \(5.0\:\mathrm{N}\). Since scales B and D show the same reading, we assume that the weight of scales A and C is distributed equally on these two (B and D) scales. The total force to be distributed among the four scales is the sum of the weight of the sack (\(220.0\mathrm{N}\)) and the weight of the scales (\(4\times5.0\:\mathrm{N}=20.0\:\mathrm{N}\)), giving a total force of \(240.0\:\mathrm{N}\). Let \(F_{B}\) and \(F_{D}\) represent the reading of scales B and D, and \(F_{A}\) and \(F_{C}\) the reading of scales A and C. Since scales B and D support the weight of scales A and C, we have: \(F_{A}=F_{C}=5.0\:\mathrm{N}\). The remaining force to be supported by scales B and D is \(240.0\:\mathrm{N} - 2\times5.0\:\mathrm{N} = 230.0\:\mathrm{N}\). Since scales B and D show the same reading, the force on each scale (F) is: \(F_{B}=F_{D}=\frac{230.0\:\mathrm{N}}{2}=115.0\:\mathrm{N}\). So the reading of each scale under the given conditions is: Scale A: \(5.0\:\mathrm{N}\) Scale B: \(115.0\:\mathrm{N}\) Scale C: \(5.0\:\mathrm{N}\) Scale D: \(115.0\:\mathrm{N}\)