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The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed: will the scale still read more than your weight at rest? Consider the following example.

What does the bathroom scale read in an elevator?

[link] shows a 75.0-kg man (weight of about 165 lb) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of 1 . 20 m/s 2 size 12{1 "." "20 m/s" rSup { size 8{2} } } {} , and (b) if the elevator moves upward at a constant speed of 1 m/s.

A person is standing on a bathroom scale in an elevator. His weight w is shown by an arrow pointing downward. F sub s is the force of the scale on the person, shown by a vector starting from his feet pointing vertically upward. W sub s is the weight of the scale pointing vertically downward. W sub e is the weight of the elevator, shown by the broken arrow pointing vertically downward. F sub p is the force of the person on the scale, acting vertically downward. F sub t is the force of the scale on the floor of the elevator, pointing vertically downward, and N is the normal force of the floor on the scale, pointing upward. (b) The same person is shown on the scale in the elevator, but only a few forces are shown acting on the person, which is our system of interest. W is shown by an arrow acting downward, and F sub s is the force of the scale on the person, shown by a vector starting from his feet pointing vertically upward. The free-body diagram is also shown, with two forces acting on a point. F sub s acts vertically upward, and w acts vertically downward.
(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. T size 12{T} is the tension in the supporting cable, w size 12{w} is the weight of the person, w s size 12{w rSub { size 8{s} } } {} is the weight of the scale, w e size 12{w rSub { size 8{e} } } {} is the weight of the elevator, F s size 12{F rSub { size 8{s} } } {} is the force of the scale on the person, F p size 12{F rSub { size 8{p} } } {} is the force of the person on the scale, F t size 12{F rSub { size 8{t} } } {} is the force of the scale on the floor of the elevator, and N size 12{N} is 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 equal F p size 12{F rSub { size 8{p} } } {} , the magnitude of the force the person exerts downward on it. [link] (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 in [link] (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 weight w size 12{w} {} and the upward force of the scale F s size 12{F rSub { size 8{s} } } {} . According to Newton’s third law F p size 12{F rSub { size 8{p} } } {} and F s size 12{F rSub { size 8{s} } } {} are equal in magnitude and opposite in direction, so that we need to find F s size 12{F rSub { size 8{s} } } {} in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

F net = ma size 12{F rSub { size 8{"net"} } = ital "ma"} {} .

From the free-body diagram we see that F net = F s w size 12{F rSub { size 8{"net"} } =F rSub { size 8{s} } - w} {} , so that

F s w = ma size 12{F rSub { size 8{s} } - w= ital "ma"} {} .

Solving for F s size 12{F rSub { size 8{s} } } {} gives an equation with only one unknown:

F s = ma + w size 12{F rSub { size 8{s} } = ital "ma"+w} {} ,

or, because w = mg , simply

F s = ma + mg size 12{F rSub { size 8{s} } = ital "ma"+ ital "mg"} {} .

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 size 12{a=1 "." "20"" m/s" rSup { size 8{2} } } {} , so that

F s = ( 75 . 0 kg ) ( 1 . 20 m/s 2 ) + ( 75 . 0 kg ) ( 9 . 80 m/s 2 ) size 12{F rSub { size 8{s} } = \( "75" "." "0 kg" \) \( 1 "." "20 m/s" rSup { size 8{2} } \) + \( "75" "." "0 kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) } {} ,

yielding

F s = 8 25 N size 12{F rSub { size 8{s} } =8"25 N"} {} .

Discussion for (a)

This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

F net = ma = 0 = F s w F s = w = mg F s = ( 75.0 kg ) ( 9. 80 m/s 2 ) F s = 735 N. alignl { stack { size 12{F rSub { size 8{"net"} } = ital "ma"=0=F rSub { size 8{s} } - w} {} #F rSub { size 8{s} } =w= ital "mg" {} # F rSub { size 8{s} } = \( "75" "." 0" kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) {} #F rSub { size 8{s} } ="735"" N" "." {} } } {}

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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