A 2.0 -m-long rod has density \(\lambda\) in kilograms per meter of length described by \(\lambda=a+b x,\) where \(a=1.0 \mathrm{kg} / \mathrm{m}, b=1.0 \mathrm{kg} / \mathrm{m}^{2}\) and \(x\) is the distance from the left end of the rod. The rod rests horizontally with each end supported by a scale. What do the two scales read?

First, calculate the total mass of the rod by integrating the given density function: \(\lambda=a+bx\), over the length of the rod (0 to L): \[M = \int_0^L \lambda dx = \int_0^2 (1.0 + 1.0x) dx\]

Evaluate the integral to calculate the total mass: \[M = [x + 0.5x^2]_0^2 = [2 + 0.5(2)^2] - [0 + 0.5(0)^2] = 4 kg\]

Since the sum of the scale readings must equal to the weight of the rod (which is \(mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity), each scale's reading can be calculated by considering the position of the mass that it carries. The center of mass for a uniform rod lies in its geometric center, but the additional mass due to the increasing density lies on the right half of the rod, closer to the right end. Therefore, the scale on the right will carry more weight than the one on the left. Since the exact distribution is hard to determine from the function, we will treat the uniform rod and the part with additional mass separately. The total weight of the rod is \(mg = 4kg * 9.8m/s^2 = 39.2N\). Half of this weight will be evenly distributed on each side due to the symmetrical uniform rod. Thus, each scale will carry at least half of this weight, or \(19.6N\). The additional mass due to the changing density is more towards the right end, hence, the right scale will read: \((3/2) * 4kg*9.8m/s^2 = 58.8N\) and the left scale will read: \(39.2N - 58.8N = -19.6N\)