Name: lawn roller force calculation
Uploaded: 2023-09-22T11:49:31.000Z
Duration: 1 min 17 s
Channel: JoVE

Рассмотрим газонный каток массой 100 кг и радиусом 25 см. Рычаг ролика выдвигается под углом тридцать градусов к горизонтальной оси.
Пренебрегая трением, развиваемым на оси, какая сила требуется для перемещения ролика с постоянной скоростью, при условии, что коэффициент сопротивления качению составляет 25 мм?
Предполагая, что результирующая движущая сила, действующая на рукоятку, прикладывается вдоль рычага, строится схема ролика в свободном теле с указанием веса, нормальной силы и движущей силы.
Движущая сила может быть разложена на горизонтальную и вертикальную составляющие.
Геометрия системы может быть использована для определения угла между нормальной силой и вертикальной осью.
Условие мгновенного равновесия затем применяется к точке контакта O. Путем подстановки значений веса, составляющих движущей силы и соответствующих перпендикулярных расстояний, оценивается движущая сила.
Если приложена сила с величиной намного меньше движущей силы, ролик не двигается.

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	<li>Хиббелер, Р.К. (2016). Инженерная механика \u2012 Статика и динамика. Хобокен, Нью-Джерси: Пирсон Прентис Холл. С. 452 \u2012 453.</li>
	<li>Мериам, Дж.Л.; Крайге, Л.Г. и Болтон, Д.Н. (2020). Инженерная механика \u2012 Статика. Хобокен, Нью-Джерси: Джон Уайли. С. 373.</li>
	<li>Бир, Ф..; Джонстон, Э.Р.; Мазурек, Д.Ф; Кромвель,.Д. и Селф, Б.. (2019). Векторная механика для инженеров \u2012 Статика и динамика. Нью-Йорк: Макгроу-Хилл. С. 463</li>
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Сопротивление скольжению: решение задачи

Сопротивление скольжению, также известное как скольжение трения, является силой, которая препятствует движению катающегося объекта, такого как колесо, шина или мяч, по поверхности. Оно вызывается деформацией объекта и поверхности, находящейся в контакте, а также другими факторами, такими как внутреннее трение, гистерезис и потери энергии в материалах. Сопротивление скольжению противодействует движению объекта, требуя дополнительной энергии для его преодоления и поддержания движения. В практическом применении, минимизация сопротивления скольжению может повысить энергоэффективность и уменьшить износ как самого катающегося объекта, так и поверхности.

Рассмотрим газонный каток с массой 100 кг, радиусом 25 см и коэффициентом сопротивления скольжению 25 мм. Рычаг катка расположен под углом 30º к горизонтальной оси. Силу, необходимую для перемещения катка с постоянной скоростью, можно определить, если игнорировать трение, возникающее на оси.

Предполагая, что результирующая сила, действующая на ручку, приложена вдоль рычага, можно построить силовую диаграмму катка, включающую вес (W), нормальную силу (N) и действующую силу (F). Действующую силу можно разложить на горизонтальные (Fh) и вертикальные (Fv) компоненты.

<img alt="Figure 1" src="/files/ftp_upload/14329/14329_PT_Fig_1v2.svg" style="text-align: center; height: 200px;" />

Затем геометрия системы может быть использована для определения угла θ1 между нормальной силой (N) и вертикальной осью. Затем применяется условие равновесия момента к точке контакта O.

<img alt="Equation 1" src="/files/ftp_upload/14329/14329_PT_Equation_1v2.svg" style="text-align: center; height: 15px;" />

Вес катка, который равен произведению массы на ускорение свободного падения, составляет 98,1 Н. Подставляя значения веса, горизонтальных и вертикальных компонентов действующей силы и соответствующие перпендикулярные расстояния, определяется действующая сила F равной 120,86 Н.

Чтобы поддерживать постоянную скорость, горизонтальная компонента действующей силы должна преодолеть сопротивление скольжению. Если приложить силу значительно меньшую по величине, чем действующая сила, каток не будет двигаться. Понимание этой зависимости помогает оценить влияние различных сил на движение катка и важность учета различных факторов при проектировании и использовании газонных катков.

Сопротивление качению, также известное как трение качения, — это сила, которая сопротивляется движению движущегося объекта, такого как колесо, шина или мяч, когда он ...

Education

JoVE Core

Mechanical Engineering

Машиностроение

Unit 1

Раздел 1

Friction

Трение

KEY TERMS AND CONCEPTS

ОСНОВНЫЕ ТЕРМИНЫ И ПОНЯТИЯ

Rolling Resistance: Problem Solving

lawn roller force calculation

Rolling resistance is the force that makes a rolling object harder to move across a surface. It is also called rolling friction. It comes from deformation in the object and the surface that touch each other. Internal friction, hysteresis, and other energy losses in the materials also play a role. Because of this force, extra energy is needed to keep a wheel, tire, or ball moving.

The video applies this idea to a lawn roller. The roller has a mass of 100 kg, a radius of 25 cm, and a coefficient of rolling resistance of 25 mm. The roller arm is set at a 30º angle to the horizontal axis. To find the force needed to move the roller at constant speed, axle friction is ignored. The driving force is assumed to act along the arm.

A free-body diagram is used to show the forces on the roller. These include the weight, the normal force, and the driving force. The driving force is split into horizontal and vertical parts. The geometry of the setup is then used to find the angle θ1 between the normal force and the vertical axis. Next, moment equilibrium is applied at the contact point O.

The weight of the roller is found by multiplying mass by gravitational acceleration. This gives 98.1 N. After the weight, the force components, and the perpendicular distances are substituted into the equilibrium equation, the driving force is calculated as 120.86 N. To keep the roller moving at a constant speed, the horizontal part of the driving force must overcome rolling resistance. If the applied force is much smaller than this value, the roller does not move. This makes the problem useful for understanding how force balance affects the motion of a lawn roller and why rolling resistance matters in design and use.

Lawn Roller Force Calculation

dry friction

Dry Friction in Crates and Everyday Motion

Dry friction appears when two solid surfaces touch and try to move past each other. Students can see it when they walk, slide a box across a table, or rub their hands together. Even though it is common, the forces behind it are not easy to see. A simple crate on a rough horizontal floor helps explain the idea. When an external force pushes the crate sideways, two main forces act on it. The normal force acts perpendicular to the surface and helps balance the crate’s weight, while the frictional...

static friction

Static Friction at the Point of Slip

Static friction is the force that keeps two surfaces from starting to slide past each other. It acts when there is a tendency for motion, but not yet actual slipping. This force helps explain everyday actions like walking on the ground and driving a car. A towing example shows how static friction changes as force is applied. A truck pulls a car with a rope, and the car stays still at first. The car is in static equilibrium, which means the forces on it are balanced. Gravity and the normal...

kinetic friction

Sliding Friction Between Moving Surfaces

Kinetic friction is the sliding friction that acts when two surfaces move against each other. It opposes relative motion and acts parallel to the contact surface, in the opposite direction of travel. In the truck and car example, the car first stays still because static friction holds it in place. When the truck’s force becomes greater than the limiting static frictional force, the friction at the contact surface drops to a lower value. The car then begins to slide, and its speed increases. At...

characteristics of dry friction

Static and Kinetic Dry Friction

Dry friction is the resistance between two solid surfaces that move against each other without lubrication or any fluid. A common example is a gardener pushing a wheelbarrow across the ground. The wheel and the ground create dry friction that slows the motion. Before the wheelbarrow starts to move, static friction acts at the contact surface. Static friction is the force that opposes motion before slipping begins, and it helps keep the wheelbarrow stable. It is present only when there is no...

types of friction problems

Dry Friction in Crate, Ladder, and Frame Problems

Dry friction problems often focus on crates, ladders, and frames that are close to slipping. Dry friction is the force that resists relative motion, or the tendency for motion, between two surfaces in contact. In many physics and engineering problems, this force helps keep an object still or defines the point where motion is about to begin. One common case has no clear impending motion. A force may be applied to a crate, but the crate stays at rest. The friction force between the floor and the...

friction: problem solving

Friction on a Banked Road

Friction on a banked road affects whether a bus stays steady while it moves at constant speed. In this problem, a bus with a mass of three megagrams travels on a sloped road. The coefficient of static friction between the tires and the road is 0.5. The goal is to find the maximum bank angle where the bus will not slip or tip. To solve the motion part, the forces on the bus must be shown first. These include gravity, the normal force, and friction. The normal force is the push from the road,...

wedges

Wedges and Self-Locking Stability

A wedge is a simple machine that can lift, support, and split heavy objects. It can also change the height of structural or mechanical parts. Because it amplifies force, a wedge makes it easier to move or hold large loads with less effort. One example is lifting a heavy slab. In that case, the wedge turns the applied force into a much larger force that acts almost perpendicular to the original push. That force change helps raise the load without requiring a large input force. To analyze wedge...

frictional forces on screws

Square-Thread Screw Mechanics and Friction

Square-threaded screws use a helical thread to turn force into lifting motion. They are common fasteners, but they are also useful for transmitting power and motion in machines. A jack with a square-threaded screw lifts heavy loads when force is applied at the handle and the screw turns. A screw can be viewed as a cylindrical shaft with an inclined square ridge wrapped around it. The distance between two neighboring threads is called the pitch. The lead is the distance the screw advances in...

upward impending motion

Screw Jack Force and Friction Analysis

A square-threaded screw jack lifts heavy loads by turning handle force into a twisting moment. Its upward motion depends on overcoming static friction between the screw threads and the jack. To study this motion, the thread can be treated as if it were a block on the groove of the jack after the thread is completely unraveled. A free-body diagram helps show the forces acting on this simplified system. The force at the handle creates a horizontal force that pushes the movable thread up the...

self-locking screw

Screw Jack Self-Locking Behavior

A square-threaded screw jack can hold a heavy load because of its self-locking behavior. This feature makes the device more reliable when it is used to lift large weights or apply a strong force. If the screw keeps its position after the turning moment is removed, the jack is self-locking. Self-locking happens because friction changes the way the forces act on the thread. The friction force reverses direction, and the reaction force moves to the other side of the thread’s normal. In this case,...

screw: problem solving

Screw Torque and Self-Locking

Screw torque and self-locking are analyzed by looking at a threaded screw shaft and a plate gear. The shaft has a mean radius and a lead. The plate gear also has a mean radius. The coefficient of static friction between the screw and the gear is part of the setup. The first step is to find the static friction angle, φ. This angle is defined by the coefficient of static friction. Next, the lead angle is calculated from the lead and the mean radius of the shaft. It gives the screw geometry...

frictional forces on flat belts

Flat Belt Tension and Friction

Flat belt tension and friction control how power moves from one pulley to another. Flat belts are used in many industrial systems to transmit power between pulleys. As the belt wraps around the pulleys, the tension changes because of friction between the belt and the pulley surface. When a pulley turns counterclockwise, the belt side moving away from the pulley has the higher tension, called T2. The side moving toward the pulley has the lower tension, called T1. To estimate these tensions, the...

flat belts: problem solving

Flat Belt Forces and Pulley Moment

Flat belts transmit power from one pulley to another, and the forces on the belt help determine the moment on a pulley. In this example, a belt wraps around pulleys A and B, with radii of 30 cm and 10 cm. The belt meets the pulleys at a 20 degree angle from the horizontal, which sets up the force balance in the system. As pulley B rotates clockwise, it drives pulley A. That motion creates tension T2 on one side of the belt and tension T1 on the other side. The belt-to-surface contact angle, β,...

pivot bearings

Axial Load Support in Pivot Bearings

Pivot bearings support axial loads on a rotating shaft. A bearing helps two parts move relative to each other while reducing friction and wear. In this design, the pivot is placed at the end of the shaft to carry the axial thrust. The pivot surface can be flat, conical, or truncated, depending on the application. In a flat pivot bearing, the axial load is spread over the bearing area. The pressure is uniform and can be found by dividing the load by the area. To study the forces more closely,...

collar bearings

Friction Torque in Collar Bearings

Collar bearings support axial loads on rotating shafts, and their friction helps determine the torque needed to start motion. They are used in many machines and may have a single collar or multiple collars, depending on the application and load needs. For a single collar bearing under an axial load, the contact area is the ring-shaped region between the external radius and the internal radius of the collar. If the bearing is assumed to support the load evenly, the uniform normal pressure is...

journal bearings

Friction Moment in Journal Bearings

Journal bearings support rotating shafts and axles while helping keep them stable. They are used in machines such as engines, turbines, and pumps. Their job is to reduce friction, wear, and vibration. A journal bearing works by keeping a thin film of lubricant between the bearing surface and the rotating shaft. This film limits direct contact and lowers frictional force. The video uses a rope winch with dry or only partly lubricated journal bearings to show the idea. As the spool turns...

bearings: problem solving

Double-Collar Bearing Torque Calculations

Double-collar bearing torque calculations show how friction and axial force affect bearing performance. Engineers use these calculations to test whether the bearing can handle the forces and moments it experiences during operation. This helps them choose and design bearings for specific applications and setups. For a double-collar bearing with set dimensions and an axial force applied, the maximum frictional moment can be found with a standard expression. The equation is rearranged, and the...

rolling resistance

Tire Deformation and Rolling Resistance

Tire deformation and rolling resistance explain why a rolling wheel still needs a driving force. When a solid cylinder or tire rolls on a rigid surface, the normal force is perpendicular to the surface at the contact point. But real materials are not perfectly rigid, so the surface pushes back with a spread of normal pressures instead of one single force. A hard cylinder on a softer surface compresses the material under it. As the cylinder moves forward, the material in front of the contact...

rolling resistance: problem solving

Rolling resistance is the force that makes a rolling object harder to move across a surface. It is also called rolling friction. It comes from deformation in the object and the surface that touch each other. Internal friction, hysteresis, and other energy losses in the materials also play a role. Because of this force, extra energy is needed to keep a wheel, tire, or ball moving. The video applies this idea to a lawn roller. The roller has a mass of 100 kg, a radius of 25 cm, and a coefficient...

key terms and concepts

Solve a lawn roller force problem using rolling resistance, free-body diagrams, and moment equilibrium. See how the required force is calculated step by step.

Lawn Roller Force Calculation | JoVE K12

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mechanical engineering

Lawn Roller Force Calculation

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