Question

Four identical spring scales, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) are used to hang a \(220.0-\mathrm{N}\) sack of potatoes. (a) Assume the scales have negligible weights and all four scales show the same reading. What is the reading of each scale? (b) Suppose that each scale has a weight of \(5.0 \mathrm{N} .\) If scales \(\mathrm{B}\) and \(\mathrm{D}\) show the same reading, what is the reading of each scale?

Short answer

Answer: In case (a), when the scales have negligible weight, the reading of each scale is \(55.0\:\mathrm{N}\). In case (b), when each scale has a weight of \(5.0\:\mathrm{N}\), the readings are as follows: - Scale A: \(5.0\:\mathrm{N}\) - Scale B: \(115.0\:\mathrm{N}\) - Scale C: \(5.0\:\mathrm{N}\) - Scale D: \(115.0\:\mathrm{N}\)

Step-by-step solution

Firstcase (a) - Negligible weight scales

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: \(F=\frac{220.0\:\mathrm{N}}{4}=55.0\:\mathrm{N}\).

Secondcase (b) - Weighted scales

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: \(115.0\:\mathrm{N}\)