Question

4.25 You have just joined an exclusive health club, located on the top floor of a skyscraper. You reach the facility by using an express elevator. The elevator has a precision scale installed so that members can weigh themselves before and after their workouts. A member steps into the elevator and gets on the scale before the elevator doors close. The scale shows a weight of 183.7 lb. Then the elevator accelerates upward with an acceleration of \(2.43 \mathrm{~m} / \mathrm{s}^{2},\) while the member is still standing on the scale. What is the weight shown by the scale's display while the elevator is accelerating?

Short answer

Answer: The weight shown by the scale's display while the elevator is accelerating is 229.3 lb.

Step-by-step solution

Convert the weight to mass in SI units

The weight of the person is given in lb. To calculate the person's mass, we need to convert the weight to SI units (Newtons) and then divide by the acceleration due to gravity (9.81 m/s²). To convert pounds to Newtons, use the following conversion factor: 1 lb = 4.44822 Newtons. So, the person's weight in Newtons is: \(W = 183.7 \,\text{lb} \times 4.44822 \,\frac{\text{N}}{\text{lb}} = 817.64834\, \text{N}\) Now, we will find the person's mass (m) using the equation \(W = m * g\), where g is the acceleration due to gravity (9.81 m/s²): \( m = \frac{W}{g} = \frac{817.64834\, \text{N}}{9.81\, \mathrm{m}/\mathrm{s}^2} = 83.41738\, \mathrm{kg}\)

Calculate the net force acting on the person

When the elevator accelerates upward, there will be a net force acting on the person. According to Newton's second law, the net force is given by \(F_\text{net} = m \times a\), where a is the acceleration of the elevator. Given the elevator's acceleration is 2.43 m/s², the net force is: \(F_\text{net} = 83.41738\, \mathrm{kg} \times 2.43\, \mathrm{m}/\mathrm{s}^2 = 202.7052\, \text{N}\)

Calculate the total force acting on the person

When the elevator is accelerating, the scale measures the normal force (N) acting on the person, which is the sum of the person's actual weight (W) and the net force (F_net) acting on them due to the elevator's acceleration: \(N = W + F_\text{net} = 817.64834\, \text{N} + 202.7052\, \text{N} = 1020.35354\, \text{N}\)

Convert the total force back to pounds

Now, we need to convert the total force in Newtons back to pounds to get the apparent weight shown by the scale: \(\text{Apparent Weight} = \frac{1020.35354\, \text{N}}{4.44822\, \frac{\text{N}}{\text{lb}}} = 229.3\, \text{lb}\) Therefore, the weight shown by the scale's display while the elevator is accelerating is 229.3 lb.

Key concepts

Weight Conversion

Understanding weight conversion is essential, especially when working with different measurement systems like imperial and metric. In this exercise, the member’s weight is given as 183.7 pounds (lb), which is a unit from the imperial system. To work with Newton’s laws of motion correctly, it’s often necessary to convert this weight into Newtons (N), the metric system unit.

To make this conversion, we use the factor that 1 lb equals approximately 4.44822 N. This factor allows us to convert any given weight from pounds to Newtons by simply multiplying the weight in pounds by this conversion factor. For instance, the member's weight in our problem converts to:

• \(183.7 \, \text{lb} \times 4.44822 \, \text{\frac{N}{lb}} = 817.64834 \, \text{N}\)

This conversion is the first step in applying Newton's laws since forces, such as weight, must be in Newtons to use with equations like \(F = ma\). After converting the weight, we divide by gravity (9.81 m/s²) to find the mass, a key step in many physics problems.

Apparent Weight

When the elevator moves, the force experienced by the person changes, altering their "apparent weight." Normally, weight is the gravitational force acting on a mass. However, in accelerating systems, like our elevator, the scale measures not just the gravitational pull but also the extra force due to acceleration. This is what we call the "apparent weight."

The apparent weight increases when the elevator accelerates upward because the scale must support both the gravitational pull and the additional force from acceleration. This force is captured with Newton's second law: \(F_{\text{net}} = m \times a\) where \(a\) is the acceleration. In our scenario:

• The calculated net force from the elevator's upward acceleration is \(202.7052 \, \text{N}\).
• Adding this to the gravitational force gives the total force or normal force the scale registers: \( N = W + F_{\text{net}} = 817.64834 \, \text{N} + 202.7052 \, \text{N} = 1020.35354 \, \text{N} \)

This total force is the apparent weight displayed by the scale, which in turn converts back to 229.3 lbs.

Net Force Calculation

Calculating the net force is crucial for understanding the interaction of different forces acting on an object. In this situation, Newton's second law is used. It explains the relationship between an object's mass, the acceleration it experiences, and the resultant force.

When the member stands on the scale in the elevator, two forces act on her: the gravitational force (her true weight) and the force due to the elevator's upward acceleration. The net force comes from this acceleration. Here's how it's determined:

• The member's mass, once converted to kilograms, is \(83.41738 \, \text{kg}\).
• The elevator's upward acceleration is \(2.43 \, \text{m/s}^2\).
• Applying Newton's second law, \( F_{\text{net}} = m \times a = 83.41738 \, \text{kg} \times 2.43 \, \text{m/s}^2 = 202.7052 \, \text{N} \)

The net force calculation is vital to determine the total force experienced by the person. Adding this net force to the gravitational force provides the apparent weight. Understanding this calculation helps highlight how forces combine and affect a person's perceived weight when under different accelerations.