Elevator acceleration This example also illustrates the full problem-solving process from Section 3.2, which incorporates the principles of dimensional analysis from Sections 3.3–3.5. A person with a mass of 70kg is standing on a scale in an elevator that reads 140 lb at a given instant. Determine

Question

Elevator acceleration

This example also illustrates the full problem-solving process from Section 3.2,which incorporates the principles of dimensional analysis from Sections 3.3–3.5.A person with a mass of 70kg is standing on a scale in an elevator that reads 140 lb at a given instant. Determine which way the elevator is moving and whether it is accelerating. From Newton’s second law, if a body is accelerating, then the sum of the forces equals the body’s mass m times its acceleration a through the equation

[latex]\sum{F=ma} [/latex]

If the sum of the forces in any direction is equal to zero, then the body is not accelerating in that direction. (See Figure 3.7.)

Approach

We are tasked to determine which direction the elevator is moving and whether the elevator is accelerating. We first make a number of assumptions about the system:

• The person and elevator are moving in tandem; so we only have to analyze the forces on the person
• The only movement is in the vertical or y-direction
• Our analysis is done on Earth, and therefore gravity is [latex]9.81 m/s^{2} or 32.2 ft/s^{2}[/latex]
• The scale does not move relative to the elevator floor or the person

We will first convert the mass of the person in kilograms to the equivalent weight in newtons. Then, we will convert their weight in newtons to pounds. We will compare the person’s weight to the reading on the scale to determine which direction the elevator is moving. Then we will use the difference in weight to determine the acceleration of the elevator. If there is no difference in weight, then we know that the elevator is not accelerating.

Accepted Answer

The weight W of the person can be found as follows:

[latex]W = (mass)(gravity) = (70 kg)(9.81 m/s^{2}) = 687 N[/latex]

Then, the weight can be converted into pounds using the factor from
Table 3.6

Table 3.6 Conversion Factors between Certain Quantities in the USCS and SI

Quantity	Conversion
Length	[latex]1 in. = 25.4 mm[/latex]
	[latex]1 in. = 0.0254 m[/latex]
	[latex]1 ft = 0.3048 m[/latex]
	[latex]1 mi = 1.609 km[/latex]
	[latex]1 mm = 3.9370 ×10^{-2} in.[/latex]
	[latex]1 m = 39.37 in.[/latex]
	[latex]1 m = 3.2808 ft[/latex]
	[latex]1 km = 0.6214 mi[/latex]
Area	[latex]1 in^2 = 645.16 mm^2[/latex]
	[latex]1 ft^2 = 9.2903 × 10^{-2} m^2[/latex]
	[latex]1 mm^2 = 1.5500 × 10^{-3 }in^2[/latex]
	[latex]1 m^2 = 10.7639 ft^2[/latex]
Volume	[latex]1 ft^3 = 2.832 × 10^{-2} m^3[/latex]
	[latex]1 ft^3 = 28.32 L[/latex]
	[latex]1 gal = 3.7854 × 10^{-3} m^3[/latex]
	[latex]1 gal = 3.7854 L[/latex]
	[latex]1 m^3 = 35.32 ft^3[/latex]
	[latex]1 L = 3.532 × 10^{-2} ft^3[/latex]
	[latex]1 m^3 = 264.2 gal[/latex]
	[latex]1 L = 0.2642 gal[/latex]
Mass	[latex]1 slug = 14.5939 kg[/latex]
	[latex]1 lbm = 0.45359 kg[/latex]
	[latex]1 kg = 6.8522 × 10^{-2} slugs[/latex]
	[latex]1 kg = 2.2046 lbm[/latex]
Force	1 lb = 4.4482 N
Force	1 N = 0.22481 lb
Pressure or stress	[latex]1 psi = 6895 Pa[/latex]
	[latex]1 psi = 6.895 kPa[/latex]
	[latex]1 Pa = 1.450 × 10^{-4 }psi[/latex]
	[latex]1 kPa = 0.1450 psi[/latex]
Work, energy, or heat	1 ft · lb = 1.356 J
	1 Btu = 1055 J
	1 J = 0.7376 ft · lb
	[latex]1 J = 9.478 × 10^{-4} Btu[/latex]
Power	1 (ft · lb)/s = 1.356 W
	1 hp = 0.7457 kW
	1 W = 0.7376 (ft · lb)/s
	1 kW = 1.341 hp

as follows,

[latex]W=687 N\left(0.22481\frac{lb}{N} \right) [/latex]

[latex]=154 \cancel{N}.\frac{lb}{\cancel{N}}[/latex]

[latex] = 154 lb[/latex]

This is the downward force of the person, represented by [latex]F_{weight}[/latex] in Figure 3.7. The scale reading represents the upward force exerted on the person by the scale, or the normal force, [latex] F_{normal}[/latex]. Since the weight of the person is larger than the scale reading, the elevator is accelerating downward which decreases the scale reading. This is illustrated in the equation

[latex]\sum{F}= F_{normal}-F_{weight}= 140 lb - 154 lb = 214 lb [/latex]

Finally, we solve for the acceleration, noting that the mass of the person must be converted to slugs.

[latex]a=\frac{\sum{F} }{m} =\left(\frac{-14 lb}{154 lbm} \right)\left(\frac{32.2 lbm}{1 slug} \right)\left(\frac{slug.\frac{ft}{s^{2} } }{lb} \right) =-2.9\frac{ft}{s^2} [/latex]

Discussion
First, we evaluate the order of magnitude of the solution. The acceleration is not large, which is not expected since the scale reading is not significantly different from the person’s weight. Second, we revisit our assumptions to make sure they are reasonable. All the assumptions are very logical. The person, scale, and elevator may undergo some motion relative to one other in reality, but the impact on the analysis would be minimal. Third, we draw conclusions from the solution and explain its physical meaning. The acceleration is negative, indicating a downward acceleration, which aligns with the scale reading. When an elevator begins to accelerate downward, the passengers temporarily feel lighter. Their mass does not change since gravity did not change. However, their perceived weight has changed, which is what the scale measures.

Note that the same analysis can be done using the SI. First, we convert the scale reading to newtons

[latex]W=(140 lb)(\frac{4.45 N}{1 lb} ) [/latex]

[latex] = 623\cancel{lb}.\frac{N}{\cancel{lb}} =623 N[/latex]

Since this is less than the person’s actual weight of 687 N, we conclude
that the elevator is accelerating downward. Solving for acceleration gives

[latex]a=\frac{\sum{F} }{m}=(\frac{623 N-687 N}{70 Kg}) [/latex]

[latex]=\frac{-64 N}{70 Kg} =-0.91(\frac{Kg.\frac{m}{s^2} }{Kg} )=-0.91 \frac{m}{s^2} [/latex]

Using the conversion [latex]1 ft = 0.3048 m[/latex] from Table 3.6, this solution can be converted to [latex]-2.9 ft/s^2[/latex], which matches our previous analysis.

[latex]a=-2.9\frac{ft}{s^2} [/latex]

Question 3.7: Elevator acceleration This example also illustrates the full...

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