College Physics

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Figure 4.25(a) The various forces acting when a person stands on a bathroom scale in an elevator. The arrows are approximately correct for when the elevator is

accelerating upward—broken arrows represent forces too large to be drawn to scale.Tis the tension in the supporting cable,wis the weight of the person,wsis


the weight of the scale,weis the weight of the elevator,Fsis the force of the scale on the person,Fpis the force of the person on the scale,Ftis the force of the


scale on the floor of the elevator, andNis the force of the floor upward on the scale. (b) The free-body diagram shows only the external forces acting on the designated


system of interest—the person.

Strategy

If the scale is accurate, its reading will equalFp, the magnitude of the force the person exerts downward on it.Figure 4.25(a) 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 inFigure 4.25(b). Analysis of the free-body diagram using Newton’s
laws can produce answers to both parts (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on

the person are his weightwand the upward force of the scaleFs. According to Newton’s third lawFpandFsare equal in magnitude and


opposite in direction, so that we need to findFsin order to find what the scale reads. We can do this, as usual, by applying Newton’s second


law,

Fnet=ma. (4.77)


From the free-body diagram we see thatFnet=Fs−w, so that


Fs−w=ma. (4.78)


Solving forFsgives an equation with only one unknown:


Fs=ma+w, (4.79)


or, becausew=mg, simply


Fs=ma+mg. (4.80)


No assumptions were made about the acceleration, and so this solution should be valid for a variety of accelerations in addition to the ones in
this exercise.
Solution for (a)

In this part of the problem,a= 1.20 m/s^2 , so that


F (4.81)


s= (75.0 kg)(1.20 m/s


2 )+ (75.0 kg)(9.80 m/s 2 ),


yielding

150 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION


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