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Chapter 4: Problem 25

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

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

Step by step solution

01

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): \( m = \frac{W}{g} = \frac{817.64834\, \text{N}}{9.81\, \mathrm{m}/\mathrm{s}^2} = 83.41738\, \mathrm{kg}\)
02

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: \(F_\text{net} = 83.41738\, \mathrm{kg} \times 2.43\, \mathrm{m}/\mathrm{s}^2 = 202.7052\, \text{N}\)
03

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: \(N = W + F_\text{net} = 817.64834\, \text{N} + 202.7052\, \text{N} = 1020.35354\, \text{N}\)
04

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.

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Most popular questions from this chapter

In a physics laboratory class, three massless ropes are tied together at a point. A pulling force is applied along each rope: \(F_{1}=150 . \mathrm{N}\) at \(60.0^{\circ}, F_{2}=200 . \mathrm{N}\) at \(100 .^{\circ}, F_{3}=100 . \mathrm{N}\) at \(190 .^{\circ} .\) What is the magnitude of a fourth force and the angle at which it acts to keep the point at the center of the system stationary? (All angles are measured from the positive \(x\) -axis.)

A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

-4.45 A pinata of mass \(M=8.0 \mathrm{~kg}\) is attached to a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.0 \mathrm{~m},\) and the top of the right pole is a vertical distance \(h=0.50 \mathrm{~m}\) higher than the top of the left pole. The pinata is attached to the rope at a horizontal position halfway between the two poles and at a vertical distance \(s=1.0 \mathrm{~m}\) below the top of the left pole. Find the tension in each part of the rope due to the weight of the pinata.

An elevator cabin has a mass of \(358.1 \mathrm{~kg},\) and the combined mass of the people inside the cabin is \(169.2 \mathrm{~kg} .\) The cabin is pulled upward by a cable, with a constant acceleration of \(4.11 \mathrm{~m} / \mathrm{s}^{2}\). What is the tension in the cable?

On the bunny hill at a ski resort, a towrope pulls the skiers up the hill with constant speed of \(1.74 \mathrm{~m} / \mathrm{s}\). The slope of the hill is \(12.4^{\circ}\) with respect to the horizontal. A child is being pulled up the hill. The coefficients of static and kinetic friction between the child's skis and the snow are 0.152 and 0.104 , respectively, and the child's mass is \(62.4 \mathrm{~kg}\), including the clothing and equipment. What is the force with which the towrope has to pull on the child?
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