/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A 35-kg crate rests on a horizon... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 38

A 35-kg crate rests on a horizontal floor, and a 65-kg person is standing on the crate. Determine the magnitude of the normal force that (a) the floor exerts on the crate and (b) the crate exerts on the person.

Short Answer

Expert verified
(a) 981 N, (b) 637.65 N

Step by step solution

01

Calculate Total Weight on the Crate

First, determine the total weight pressing down on the crate. This accounts for both the crate's weight and the person's weight. The weight of the crate is given by the equation \( W_{crate} = m_{crate} \times g \), where \( m_{crate} = 35 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \), which makes \( W_{crate} = 35 \times 9.81 = 343.35 \text{ N} \). Similarly, calculate the weight of the person as \( W_{person} = m_{person} \times g = 65 \times 9.81 = 637.65 \text{ N} \). The total weight on the crate is then \( 343.35 + 637.65 = 981 \text{ N} \).
02

Determine the Normal Force Exerted by the Floor

The normal force that the floor exerts on the crate must be equal and opposite to the total weight acting downward, which includes both the crate and the person standing on it. Therefore, the normal force exerted by the floor on the crate is \( 981 \text{ N} \).
03

Identify the Normal Force Exerted by the Crate on the Person

The normal force exerted by the crate on the person only balances the weight of the person, since it's the force the crate applies upward to support the person. Hence, it equals the weight of the person, which is \( 637.65 \text{ N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weight Calculation
When dealing with physics problems involving forces, it's essential to calculate the weight of the objects involved. Weight is a measure of the force exerted by gravity on an object and is calculated using the formula:
  • \( W = m \times g \)
  • Where \( W \) is the weight, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \) on Earth.

In the exercise, we started by calculating the weight of the crate and the person. The crate's weight is computed as \( 35 \times 9.81 = 343.35 \text{ N} \), and the person's weight as \( 65 \times 9.81 = 637.65 \text{ N} \).
Together, their total weight pressing down on the crate is \( 343.35 + 637.65 = 981 \text{ N} \). This combined weight helps determine how much force the floor must exert upwards to support both the crate and the person.
Newton's Third Law
Newton's Third Law of Motion is a fundamental principle that states: "For every action, there is an equal and opposite reaction." This means that forces always occur in pairs.
When one object exerts a force on another, the second object exerts a force of equal magnitude but in the opposite direction back on the first object.
  • For example, when the crate and the person press down on the floor due to their combined weight, the floor responds by exerting an equal and opposite normal force upwards.
  • Thus, in the exercise, the floor exerts a force of \( 981 \text{ N} \) upwards on the crate, perfectly balancing the downward force of the crate and person.
Understanding this law is crucial when analyzing scenarios involving normal forces and ensures that all forces in the system are balanced, keeping the objects at rest or in uniform motion.
Free Body Diagram
A free body diagram is a graphical illustration used to visualize the forces acting on an object. It helps break down complex problems into simpler parts by isolating a single object and depicting all the external forces acting upon it.
  • For the problem at hand, envisioning a free body diagram can clarify how the forces work together.
  • Consider the crate on the floor, the arrows pointing downwards represent the gravitational forces due to the weights of the crate (343.35 N) and the person (637.65 N).
  • Another arrow, pointing upwards, displays the normal force exerted by the floor (981 N).

For the person standing on the crate, a separate free body diagram would show their weight, 637.65 N, acting downwards, balanced by an upward-directed normal force from the crate of the same magnitude.
These diagrams are valuable tools for identifying and confirming that forces are balanced, following Newton's laws without ambiguities.

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Most popular questions from this chapter

A 1.40-kg bottle of vintage wine is lying horizontally in a rack, as shown in the drawing. The two surfaces on which the bottle rests are \(90.0^{\circ}\) apart, and the right surface makes an angle of \(45.0^{\circ}\) with respect to the horizontal. Each surface exerts a force on the bottle that is perpendicular to the surface. Both forces have the same magnitude \(F .\) Find the value of \(F\) .

A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earth and the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m},\) and the mass of the earth is 81.4 times as great as that of the moon.

ssm A 95.0 -kg person stands on a scale in an elevator. What is the apparent weight when the elevator is \(\quad\) (a) accelerating upward with an acceleration of \(1.80 \mathrm{m} / \mathrm{s}^{2}, \quad\) (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.30 \(\mathrm{m} / \mathrm{s}^{2} ?\)

On earth, two parts of a space probe weigh 11000 \(\mathrm{N}\) and 3400 \(\mathrm{N}\) . These parts are separated by a center-to-center distance of 12 \(\mathrm{m}\) and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

At an instant when a soccer ball is in contact with the foot of a player kicking it, the horizontal or \(x\) component of the ball's acceleration is 810 \(\mathrm{m} / \mathrm{s}^{2}\) and the vertical or \(y\) component of its acceleration is 1100 \(\mathrm{m} / \mathrm{s}^{2}\) . The ball's mass is 0.43 \(\mathrm{kg}\) . What is the magnitude of the net force acting on the soccer ball at this instant?
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