Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The blades of ice skates are quite thin, so the pressure exerted on ice by a skater can be substantial. Explain how this facilitates skating on ice.

Short Answer

Expert verified
Skating is facilitated by thin blades which increase pressure, melting ice to create a slick surface.

Step by step solution

01

Understanding Pressure

Pressure is the force exerted per unit area and is calculated using the formula: \( P = \frac{F}{A} \), where \( P \) is pressure, \( F \) is the force applied, and \( A \) is the area over which the force is distributed.
02

Considering Thin Blades

The blades of ice skates are very thin, which means the area \( A \) in contact with the ice is quite small. This results in an increase in the pressure applied to the ice because the denominator (area \( A \)) is smaller.
03

Effect of Increased Pressure on Ice

When pressure is increased on the ice by the skates, it lowers the melting point of the ice under the blades. This causes a thin layer of water to form, creating a lubricating layer between the ice and the blade, which reduces friction and facilitates smooth sliding.
04

How Skating is Facilitated

The thin layer of water reduces friction, allowing the skater to glide over the ice effortlessly. The skates can move smoothly with less resistance due to this thin water film.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Melting Point Depression
The concept of melting point depression is fascinating, especially in the context of ice skating. Normally, the melting point of ice is 0°C, meaning ice transitions to water at this temperature under standard atmospheric conditions. However, when pressure is applied, such as from the thin blades of ice skates, this melting point can lower. This is due to the fact that increasing pressure affects the ice structure, encouraging it to become water even below its normal melting point.

In the context of skating, when the skater stands on the ice, the high pressure from the narrow blades forces some of the ice beneath to melt. So, even when the atmospheric temperature is below freezing, a thin layer of water is created between the blade and the ice surface. This melting point depression is a key element allowing for the smooth glide experienced during skating. It’s nearly like having a natural lubricant, increasing the efficiency of movement on the ice.
Friction Reduction
Friction is a force that opposes motion, and reducing it is crucial for smooth skating. Under normal conditions, ice is slippery because its surface already has a thin layer of water. But when ice is under pressure from the thin skate blades, even more water forms, dramatically decreasing friction.

This is essential because friction can make movement difficult, requiring more effort to slide across the ice. With less friction, skaters can glide with minimal resistance, enabling quicker and more graceful movements.
  • High pressure creates a thin water layer
  • Water layer acts as a lubricant
  • Reduced friction allows smoother skating
The glide becomes almost effortless due to this reduction in friction, helping make complex movements more attainable and controllable.
Ice Skating Mechanics
Ice skating might seem mystical, but it's primarily about physics. The mechanics involve converting the skater's energy into forward motion with minimal resistance. The thin skates increase pressure on the ice, creating a layer of water that enhances glide.

Skaters push off the ice using the edges of their skates. The thin water layer helps them transition smoothly from one foot to the other. It’s crucial for maintaining balance and direction, transforming energy into controlled movement over the friction-reduced surface.
  • Initial push-off uses skate edges
  • Water layer aids transition between feet
  • Maintains speed and direction
These mechanics of movement make skating a unique showcase of applied physics, demonstrating how scientific principles can elevate sports and recreational activities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Barium metal crystallizes in a body-centered cubic lattice (the Ba atoms are at the lattice points only). The unit cell edge length is \(502 \mathrm{pm}\), and the density of the metal is \(3.50 \mathrm{~g} / \mathrm{cm}^{3}\). Using this information, calculate Avogadro's number. [Hint: First calculate the volume (in \(\mathrm{cm}^{3}\) ) occupied by \(1 \mathrm{~mole}\) of \(\mathrm{Ba}\) atoms in the unit cells. Next calculate the volume (in \(\mathrm{cm}^{3}\) ) occupied by one \(\mathrm{Ba}\) atom in the unit cell. Assume that 68 percent of the unit cell is occupied by \(\mathrm{Ba}\) atoms.

A face-centered cubic cell contains \(8 \mathrm{X}\) atoms at the corners of the cell and \(6 \mathrm{Y}\) atoms at the faces. What is the empirical formula of the solid?

How do the boiling points and melting points of water and carbon tetrachloride vary with pressure? Explain any difference in behavior of these two substances.

The boiling point and freezing point of sulfur dioxide are \(-10^{\circ} \mathrm{C}\) and \(-72.7^{\circ} \mathrm{C}\) (at \(\left.1 \mathrm{~atm}\right)\), respectively. The triple point is \(-75.5^{\circ} \mathrm{C}\) and \(1.65 \times 10^{-3} \mathrm{~atm},\) and its critical point is at \(157^{\circ} \mathrm{C}\) and 78 atm. On the basis of this information, draw a rough sketch of the phase diagram of \(\mathrm{SO}_{2}\).

Given the general properties of water and ammonia, comment on the problems that a biological system (as we know it) would have developing in an ammonia medium. $$ \begin{array}{lll} & \mathrm{H}_{2} \mathrm{O} & \mathrm{NH}_{3} \\ \hline \text { Boiling point } & 373.15 \mathrm{~K} & 239.65 \mathrm{~K} \\ \text { Melting point } & 273.15 \mathrm{~K} & 195.3 \mathrm{~K} \\ \text { Molar heat capacity } & 75.3 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} & 8.53 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \\ \text { Molar heat of vaporization } & 40.79 \mathrm{~kJ} / \mathrm{mol} & 23.3 \mathrm{~kJ} / \mathrm{mol} \\ \text { Molar heat of fusion } & 6.0 \mathrm{~kJ} / \mathrm{mol} & 5.9 \mathrm{~kJ} / \mathrm{mol} \\ \text { Viscosity } & 0.001 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2} & 0.0254 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2} \\ & & (\text { at } 240 \mathrm{~K}) \\ \text { Dipole moment } & 1.82 \mathrm{D} & 1.46 \mathrm{D} \\ \text { Phase at } 300 \mathrm{~K} & \text { Liquid } & \text { Gas } \end{array} $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free