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Chapter 11: Problem 16

A 58 -kg skier is going down a slope oriented \(35^{\circ}\) above the horizontal. The area of each ski in contact with the snow is \(0.13 \mathrm{~m}^{2}\). Determine the pressure that each ski exerts on the snow.

Short Answer

Expert verified
The pressure exerted by each ski is approximately 3572.85 Pa.

Step by step solution

01

Identify the Forces

The skier's weight acts vertically downwards due to gravity. The normal force acts perpendicular to the surface of the slope. We need to find the component of the skier's weight that acts perpendicular to the slope.
02

Calculate the Skier's Weight

First, calculate the total weight (force) of the skier using the equation: \( F = m \cdot g \), where \( m = 58 \, \text{kg} \) and the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \). \( F = 58 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 568.4 \, \text{N} \).
03

Calculate the Normal Force Component

The normal force is the component of the skier's weight perpendicular to the slope. Since the slope is inclined at \(35^{\circ}\), the normal force is given by the formula: \( F_{\text{normal}} = F \cos(\theta) \). \( F_{\text{normal}} = 568.4 \, \text{N} \times \cos(35^{\circ}) \approx 568.4 \, \text{N} \times 0.8192 \approx 465.47 \, \text{N} \).
04

Calculate the Pressure Exerted by Each Ski

Pressure is the force exerted per unit area. The area of each ski is given as \(0.13 \, \text{m}^2\). So, calculate the pressure using the formula: \( P = \frac{F_{\text{normal}}}{A} \), where \( A = 0.13 \, \text{m}^2 \).\( P = \frac{465.47 \, \text{N}}{0.13 \, \text{m}^2} \approx 3572.85 \, \text{Pa} \).

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Key Concepts

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

Forces on Inclined Planes
When dealing with inclined planes, it's important to understand how forces interact with these surfaces. An inclined plane is any flat surface tilted at an angle to the horizontal. Forces on such a surface include:
  • Gravity: Acts straight down towards the Earth's center.
  • Normal Force: Acts perpendicular to the plane's surface.
For a skier on a slope, these forces are key to staying balanced. Gravity pulls the skier downwards, but not all this force pushes the skier straight down the slope. A portion of the gravitational force acts perpendicular to the slope, which is countered by the normal force. This is calculated using trigonometry based on the plane's angle, ensuring the skier doesn't sink through but rather glides smoothly over the snow.
Normal Force Calculation
To calculate the normal force on an inclined plane, it is necessary to break down the gravitational force into two components: parallel and perpendicular to the plane. Imagine the weight of the object acting downwards. The normal force is the component that supports the weight against the slope.

The force of gravity acting on the skier can be calculated as the product of mass and the acceleration due to gravity: \( F = m \cdot g \).Then, the normal force is found using the angle of inclination:

\( F_{\text{normal}} = F \cos(\theta) \)

where \( \theta \) is the angle between the slope and the horizontal. Remember, cosine is used because the normal force is perpendicular to the slope. In our specific problem, it allows us to adjust the total weight into a component that the slope directly supports, determining how much of the skier's weight the snow has to hold.
Pressure Formula
Pressure is a measure of force distribution over an area. It plays a crucial role in scenarios like skiing, where force must be distributed to prevent sinking into snow. The formula for pressure is:\[ P = \frac{F}{A} \]where \( F \) represents the force applied and \( A \) is the area over which this force acts.

In our exercise, after finding the normal force, which is already adjusted to the plane's inclination, we apply this to the skis, each with a contact area of \(0.13 \, \text{m}^2\). By dividing the normal force by this area, we determine the pressure exerted on the snow by each ski. This pressure distribution is what keeps the skier gliding without sinking, highlighting how crucial understanding pressure is for optimizing contact surfaces during skiing.

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Most popular questions from this chapter

One of the concrete pillars that support a house is \(2.2 \mathrm{~m}\) tall and has a radius of \(0.50 \mathrm{~m}\). The density of concrete is about \(2.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Find the weight of this pillar in pounds \((1 \mathrm{~N}=0.2248 \mathrm{lb})\).

A hollow cubical box is \(0.30 \mathrm{~m}\) on an edge. This box is floating in a lake with one-third of its height beneath the surface. The walls of the box have a negligible thickness. Water is poured into the box. What is the depth of the water in the box at the instant the box begins to sink?

A person who weighs \(625 \mathrm{~N}\) is riding a 98-N mountain bike. Suppose the entire weight of the rider and bike is supported equally by the two tires. If the gauge pressure in each tire is \(7.60 \times 10^{5} \mathrm{~Pa}\), what is the area of contact between each tire and the ground?

Some researchers believe that the dinosaur Barosaurus held its head erect on a long neck, much as a giraffe does. If so, fossil remains indicate that its heart would have been about \(12 \mathrm{~m}\) below its brain. Assume that the blood has the density of water, and calculate the amount by which the blood pressure in the heart would have exceeded that in the brain. Size estimates for the single heart needed to withstand such a pressure range up to two tons. Alternatively, Barosaurus may have had a number of smaller hearts.

A hot-air balloon is accelerating upward under the influence of two forces, its weight and the buoyant force. For simplicity, consider the weight to be only that of the hot air within the balloon, thus ignoring the balloon fabric and the basket. (a) How is the weight of the hot air determined from a knowledge of its density \(\rho\) hot air and the volume \(V\) of the balloon? (b) For a given volume \(V\) of the balloon, does the buoyant force depend on the density of the hot air inside the balloon, the density of the cool air outside the balloon, or both? Provide a reason for your answer. (c) Draw a free-body diagram for the balloon, showing the forces that act on it. How is the upward acceleration of the balloon related to these forces and to its mass?
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