## Pulleys in Everyday Life

Wells, elevators, construction sites, exercise machines and belt-driven generators are all applications that use pulleys as a basic function of the machinery.

An elevator uses counter weights with pulleys to provide a lift system for heavy objects. Belt driven generators are used to provide backup power to modern-day applications such as a manufacturing factory. Military bases use belt-driven generators to provide power to the station when there is a conflict.

The military uses generators to provide power to military bases when there is no external power supply. The applications of belt-driven generators are enormous. Pulleys are also used to lift cumbersome objects in construction, such as a human being cleaning windows on a very tall building or even lifting very heavy objects used in construction.

## Mechanics Behind Belt Driven Generators

The belt generators are powered by two different pulleys moving at two various revolutions per minute, which means how many rotations a pulley can complete in a minute.

The reason why the pulleys rotate at two different RPMs is that it affects the period or the time it takes the pulleys to complete one rotation or cycle. Period and frequency have an inverse relationship, meaning the period affects the frequency, and the frequency affects the period.

Frequency is an essential concept to understand when powering specific applications, and frequency is measured in hertz. Alternators are also another form of a pulley-driven generator that is used to recharge the battery's in the vehicles that are driven today.

Many types of generators use alternating current and some use direct current. The first direct current generator was built by Michael Faraday which showed that both electricity and magnetism are a unified force called the electromagnetic force.

## Pulley Problems in Mechanics

Pulley systems are used in mechanics problems in physics. The best way to solve pulley problems in mechanics is by utilizing Newton's second law of motion and understanding Newton's third and first laws of motion.

Newton's second law states:

*F = ma*

Where, *F* is for the net force, which is the vector sum of all the forces acting on the object. m is the mass of the object, which is a scalar quantity meaning mass only has magnitude. Acceleration gives Newton's second law its vector property.

In the given examples of pulley system problems, familiarity with algebraic substitution will be required.

The most simple pulley system to solve is a primary **Atwood's machine** using algebraic substitution. Pulley systems are usually constant acceleration systems. An Atwood's machine is a single pulley system with two weights attached with one weight on each side of the pulley. The problems regarding an Atwood's machine consist of two weights of equal mass and two weights of uneven masses.

*If an Atwood's machine consists of one 50 kilogram weight to the left of the pulley and a 100 kg weight to the right of the pulley, what is the system's acceleration?*

To begin, draw a free body diagram of all the forces acting on the system, including tension.

**Object to the right of the pulley**

m_{1}g-T=m_{1}a

Where T is for tension and g is the acceleration due to gravity.

**Object to the left of the pulley**

If tension is pulling up in the positive direction therefore the tension is positive, clockwise (going with) with respect to a clockwise rotation. If the weight is pulling down in the negative direction therefore the weight is negative, counterclockwise (opposing) with respect to a clockwise rotation.

Therefore applying Newtons second law of motion:

Tension is positive, W or m_{2}g is negative as follows

T-m_{2}g = m_{2}a

Solve for tension.

T = m_{2}g+m_{2}a

Substitute into the equation of the first object.

m_{1}g-T=m_{1}a

m_{1}g - (m_{2}g+m_{2}a)=m_{1}a

m_{1}g-m_{2}g-m_{2}a=m_{1}a

m_{1}g-m_{2}g = m_{2}a+m_{1}a

Factor:

(m_{1}-m_{2})g = (m_{2}+m_{1})a

Divide and solve for acceleration.

(m_{1}-m_{2})g/(m_{2}+m_{1}) = a

Plug in 50 kilograms for second mass and 100 kg for the first mass

(100kg-50kg)9.81m/s^{2}/(50kg+100kg) = a

490.5/150 = a

3.27 m/s^{2} = a

## Graphical Analysis of the Dynamic of a Pulley System

If the pulley system was released from rest with two unequal masses and was graphed on a velocity versus time graph, it would produce a linear model, meaning it would not form a parabolic curve but a diagonal straight line starting from the origin.

The slope of this graph would produce acceleration. If the system were graphed on a position versus time graph, it would produce a parabolic curve starting from the origin if it was realized from rest. The slope of the graph of this system would produce the velocity, meaning the velocity varies throughout the pulley system's motion.

## Pulley Systems and Frictional Forces

A **pulley system with friction** is a system that interacts with some surface that has resistance, slowing the pulley system down due to frictional forces. In this cases the surface of the table is the form of resistance interacting with the pulley system, slowing the system down.

The following example problem is a pulley system with frictional forces acting on the system. The frictional force in this case is the surface of the table interacting with the block of wood.

*A 50 kg block rests on a table with a coefficient of friction between the block and the table of 0.3 on the left side of the pulley. The second block is hanging on the right side of the pulley and has a mass of 100 kg. What is the acceleration of the system?*

To solve this problem, Newton's third and second laws of motion must be applied.

Begin by drawing a free body diagram.

Treat this problem as one dimensional, not two dimensional.

Force of friction will pull to the left of object one opposing motion. The force of gravity will pull directly down, and the normal force will pull in the opposite direction of the force of gravity equal in magnitude. Tension will pull to the right in the direction of the pulley clockwise.

Object two, which is the hanging mass to the right of the pulley, will have the tension pulling up counterclockwise and the force of gravity pulling down clockwise.

If the force is opposing the motion, it will be negative, and if the force is going with motion, it will be positive.

Then, start by calculating the vector sum of all the forces acting on the first object resting on the table.

The normal force and the force of gravity cancel out according to Newton's third law of motion.

F_{k} = u_{k}F_{n}

Where F_{k} is the force of kinetic friction, meaning the objects in motion and u_{k} is the coefficient of friction and Fn is the normal force that runs perpendicular to surface at which the object is resting.

The normal force is going to be equal in magnitude to the force of gravity, so, therefore,

F_{n} = mg

Where F_{n} is the normal force and m is the mass and g is the acceleration due to gravity.

Apply Newton's second law of motion for object one to the left of the pulley.

F_{net} = ma

Friction opposes motion tension is going with a motion so, therefore,

-u_{k}F_{n} + T = m_{1}a

Next, find the vector sum of all the forces acting on object two, which is just the force of gravity pulling directly down with motion and tension opposing the motion in the counterclockwise direction.

So, therefore,

F_{g} - T = m_{2}a

Solve for tension with the first equation that was derived.

T = u_{k}F_{n} + _{} m_{1}a

Substitute tension equation into the second equation so, therefore,

Fg-u_{k}F_{n} - _{} m_{1}a = m_{2}a

Then solve for acceleration.

Fg-u_{k}F_{n} = m_{2}a+m_{1}a

Factor.

m_{2}g-u_{k}m_{1}g = (m_{2}+m_{1})a

Factor g and dived to solve for a.

g(m_{2}-u_{k}m_{1})/(m_{2}+m_{1}) = a

Plugin the values.

9.81 m/s^{2}(100kg-.3(50kg))/(100kg+50kg) = a

5.56 m/s^{2} = a

## Pulley Systems

Pulley systems are used in everyday life, anywhere from generators to lifting heavy objects. Most importantly, pulleys teach the basics of mechanics, which is vital to understanding physics. The importance of pulley systems is essential for the development of the modern industry and is very commonly used. A physics pulley is used for belt driven generators and alternators.

A belt driven generator consist of two rotating pulleys that rotate at two different RPMs, which are used to power equipment in case of a natural disaster or for general power needs. Pulleys are used in industry when working with generators for back up power.

Pulley problems in mechanics occur everywhere from calculating loads when designing or building and in elevators to calculating the tension in the belt lifting a heavy object with a pulley so the belt doesn't break. Pulley system are not only used in physics problems by are used in the modern world today for a vast amount of applications.

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