Strategy If the scale at rest is accurate, its reading equals Fp, the magnitude of the force the person exerts downward on it. Part (a) of the figure shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn, as shown in part (b) of the figure. The only forces acting on the person are his weight w and the upward force of the scale F .. According to Newton's third law, Fp and F, are equal in magnitude and opposite in direction, so that we need to find F, in order to find what the scale reads. We can do this, as usual, by applying Newton's second law, Fnet = mã. From the free-body diagram, we see that Fuet = F, - w, so we have F. - w = ma. Solving for F, gives us an equation with only one unknown: or, because w = mg, simply F, = ma+ w, F} = ma + mg. No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. (Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton's second law becomes F, - w = - ma.) Solution 1. We have @ = 1.20 m/s2, so that F. = (75.0 kg) (9.80 m/s2) + (75.0 kg) (1.20 m/s2) yielding F= = 825 N. 2. Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity-up, down, or stationary-acceleration is zero because a = At and Av = 0. Thus, F}= ma + mg = 0 + mg or Fs =(75.0kg) (9.80 m/s2), which gives F= = 735 N.