Types of Pulleys & Calculators

IntroductionPulley systems are an essential component in many mechanical systems and have been used for thousands of years to lift heavy loads and transmit power. In this article, we will explore the different types of pulley systems, their underlying physics, and their various applications.
ExplanationA pulley is a wheel attached to a shaft or axle that carries a rope, wire or general taut cable across it. It is used to support movement by transferring power, enabling the direction of motion to be changed. 
They can be coupled together to reduce the amount of force required to move or lift a load, at the cost of requiring a longer distance to move the cable through. 
Pulleys are considered to be a part of a group of fundamental mechanical devices known as simple machines.
Different Types of PulleysIn this section, we'll explore types of pulleys you might find in practice. We also present some worked examples and pulley calculators for you to quickly check the forces acting on the different types of pulleys.
Fixed Pulley SystemsA fixed pulley system is the simplest type of pulley system, where the pulley is attached to a stationary object such as a wall or a beam. This type of pulley system does not change the direction of the applied force, but it does change the magnitude of the force. The mechanical advantage of a fixed pulley system is equal to the number of ropes supporting the load.
﻿
Fixed Pulley Systems diagram
Worked Example: Fixed Pulley SystemA load with a mass (m1) of 10kg is hoisted up by a person with a force of 1000N directly downwards on a fixed pulley. Determine the acceleration of the system and the force with which the pulley acts upon its axis. Neglect friction, the mass of string and pulley. 
First Step: Draw the diagram and list the forces!
﻿
Example of Fixed Pulley System
Here we can see all the major forces acting on the system. The applied force (1000N), The tension on the rope (T1 and T2) and the gravitational force exerted by the load (mg).
﻿
Second Step: List the relevant equations!
﻿
T2−mg=m2a1000−T1=m1aT2 - mg = m_2a \\ 1000 - T1 = m_1aT2−mg=m2​a1000−T1=m1​a
For our purposes, the m1 is a stand-in for the case where a mass and gravitational force are acting as the pulling force. 
In our system, the magnitudes of the tension on the rope are equal thus simplifying our equations. 
﻿
∣T1∣=∣T2∣=∣T∣T−m2g=m2a1000−T=m1a|T1| = |T2| = |T|\\ T - m_2g = m_2a \\ 1000 - T = m_1a∣T1∣=∣T2∣=∣T∣T−m2​g=m2​a1000−T=m1​a
Third Step: Rearrange and solve for acceleration
﻿
T=m2a+m2g=1000−m1aa(m1+m2)=1000+mga=(1000−m2g)(m1+m2)T = m_2a + m_2g = 1000 - m_1a \\ a (m_1+m_2) = 1000 + mg \\
a = \frac {(1000 - m_2g) }{( m_1+m_2) }
T=m2​a+m2​g=1000−m1​aa(m1​+m2​)=1000+mga=(m1​+m2​)(1000−m2​g)​
ur m1 value can be calculated as: 
﻿
m1=1000g m_1 = \frac {1000}{g} m1​=g1000​
﻿
Now we calculate the output! 
﻿
Here's a calculation setup and ready for you to use.
Inputs ﻿﻿m_2 
:10kg​
﻿
﻿﻿Force Applied
:1000N​
﻿
Output﻿﻿a 
:8.05722521​
﻿
﻿
💬 We'd love your feedback on this template! It takes 1min

Conclusion: By applying a 1000N force downward, we can accelerate the 10kg mass upwards with an acceleration of 8.06 ms^-2 

Movable Pulley SystemsA movable pulley system, also known as a block and tackle, is similar to a fixed pulley system but the pulley is attached to the load rather than a stationary object. This type of pulley system changes the direction of the applied force and also increases the magnitude of the force. The mechanical advantage of a movable pulley system is equal to the number of ropes supporting the load multiplied by the number of sections in the pulley system.
﻿
Worked Example: Movable/Compound Pulleys
﻿
Movable/Compound Pulleys work example
A movable pulley system has two masses, M1 = 10kg and M2 = 5kg. Pulley A is fixed, and Pulley B is movable. Assuming the pulleys are massless, the string is taut and a constant tension is maintained in the system, calculate the acceleration of M1. 
﻿
First Step: Label the diagram with relevant parameters, and list free body diagrams if applicable 
﻿
free body diagrams of Movable/Compound Pulleys
Second Step: Perform analysis and list relevant equations
If we analyse pulley B, we can recognise that the total tension on the string at both sides is equal to the magnitude of the gravitational force of mass 2. 
﻿
M2ab=2T1−T2M_2a_b = 2T1 - T2M2​ab​=2T1−T2
Assuming the string is kept taut, M2 will have the same acceleration as pulley B.
﻿
aB=a2a_B = a_2aB​=a2​
Now we must analyse the change in the length of the pulley string sections. 
Increase in length left of fixed pulley A = L1 
Decrease in length to the right of A = L2 
Decrease in length to the right of B = La 
﻿
△L=L1−L2−La=L1−2L2\triangle L = L1-L2-La = L1 - 2L2△L=L1−L2−La=L1−2L2
Note: What's important here is the change in length, not the length of the string itself, a common point of confusion!
﻿
Differentiating this change in length twice gives us the resulting accelerations we are after: 
﻿
a1=2a2a_1 = 2a_2a1​=2a2​
We can calculate the net forces acting on both pulleys using our previous analysis. 
For pulley A: 
﻿
M1g−T=M1a1M_1g - T = M_1a_1M1​g−T=M1​a1​
For pulley B: 
﻿
2T−M2g=M2a22T - M_2g = M_2a_22T−M2​g=M2​a2​
﻿
Final Step: Rearrange for acceleration and calculate!

Note: a1 = 2*a2 
﻿
a1=(4M1−2M2)g4M1+M2a_1 = \frac {(4M_1-2M_2)g}{4M_1+M_2}a1​=4M1​+M2​(4M1​−2M2​)g​
Now we calculate the output!
﻿
Let's use our in-built calculation tool.
Inputs﻿﻿M1
:10kg​
﻿
Output﻿﻿a
:6.54​
﻿
﻿﻿M2
:5kg​
﻿
﻿
Final remark: For our movable pulley system with a M1 = 10kg, M2 = 5kg, M1 accelerates downwards at a rate of 6.54 metres/second^2
﻿
Compound Pulley SystemsA compound pulley system is a combination of fixed and movable pulley systems, where multiple pulleys are used to increase the mechanical advantage of the system. This type of pulley system can provide a significant increase in mechanical advantage over a simple fixed or movable pulley system.
﻿
Compound Pulley Systems free body diagrams 
See the previous example for a worked solution using both a fixed and movable pulley!
Multiple Pulley SystemsMultiple pulley systems are used when a large mechanical advantage is required, such as in cranes and hoists. This type of pulley system can provide an even larger mechanical advantage than a compound pulley system, as the number of sections in the pulley system is increased.
﻿
Multiple Pulley System
Worked Example: Multiple Pulley SystemGiven the following pulley system, find the force F required to hold a weight W as shown in the below diagram.

﻿
Multiple Pulley Systems free body diagrams 
First Step: List the tension on the string in all sections and recognise T = F.
﻿
Multiple Pulley System free body diagrams 
It can be observed that at all points on the string, there is a constant tension T. This tension force is equal to the force variable F.
﻿
Second Step: Isolate the system involving the relevant pulleys.
﻿
Multiple Pulley System free body diagram
From this diagram, we can determine that the weight force of 100N must equal 5 multiplied by the tension T if the system is to remain in equilibrium.
﻿
Third Step: Perform analysis and determine F.
﻿
∑Fy=0W=100N=5T ∴T=20N=F\sum F_y = 0 \\ W = 100N = 5T\: \\ \therefore T = 20N = F∑Fy​=0W=100N=5T∴T=20N=F
﻿
Let's determine F using our calculation tool.
Input
Output
﻿﻿W 
:100​
﻿
﻿﻿F
:20​
﻿
Final remark: Our analysis determined that it would take a 20N force F to maintain equilibrium. Therefore we can ascertain that this pulley system allows for a mechanical advantage of 5 (100/20). 

This demonstrates that for large multiple pulley systems, we can achieve the large mechanical advantages required for machines such as cranes.
﻿
Planetary Pulley SystemsA planetary pulley system is a type of pulley system that uses a planetary gear set to transmit power. This type of pulley system is commonly used in power transmission systems and is known for its high mechanical advantage and efficiency.
What's planetary gear? 
Also known as epicyclic gears, planetary gears are a set of gears that consist of two gears mounted such that the centre of one gear revolves around the centre of the gear. 
﻿
Planetary Gears [Source]
﻿
Planetary Pulley System
 Source: https://www.youtube.com/watch?v=5I6Mhu8XRxI&ab_channel=Del
Worked Example: Planetary Gear SystemA planetary gear system is used in an automatic transmission for a vehicle. Releasing or locking certain gears gives the advantage of operating the car at different speeds. 
Sources [2] and [3] show a planetary gear system. Consider the case where the ring gear R is held fixed, WR = 0, and the sun gear S is rotating at Ws = 5 rad/s. Find the angular velocity of each of the planet gears P and shaft A. 
﻿
First step: Write out all the relevant equations! 
To solve this problem we are going to utilise the following general equations: 

The velocity of a point:
﻿
v⃗=w⃗×r⃗p\vec v = \vec w \times \vec r_p v=w×rp​
The velocity of a point with respect to another point: 
﻿
V⃗b=V⃗a+w⃗×r⃗b/a\vec V_b = \vec V_a + \vec w \times \vec r_{b/a} Vb​=Va​+w×rb/a​
﻿
Second step: Apply the equations and solve for the angular velocities! 
Using our equations in step 1, we can recognise: 
﻿
va=ws×r⃗=(5)(80)=−400 mm/sv_a = w_s \times \vec r = (5)(80) = -400 \: mm/sva​=ws​×r=(5)(80)=−400mm/s
﻿
vb=va+wp×rb/a0=−400i⃗+(wpk⃗)×(80j⃗)0=−400i⃗−80wpi⃗wp=400−80=−5 rad/sv_b = v_a + w_p \times r_{b/a} \\ 0 = -400\vec i + (w_p\vec k ) \times (80\vec j)\\ 0  = -400\vec i - 80 w_p \vec i \\ w_p = \frac {400}{-80} = -5 \: rad/svb​=va​+wp​×rb/a​0=−400i+(wp​k)×(80j​)0=−400i−80wp​iwp​=−80400​=−5rad/s
﻿
We also recognise: 
﻿
vc=wa⋅r1v_c = w_a \cdot r_1 vc​=wa​⋅r1​
We can calculate vc by using our earlier equation format: 
﻿
vc=vb+wp×rc/b=0+(−5k⃗)×(−40j⃗)vc=−200i⃗ mm/sv_c  = v_b + w_p \times r_{c/b} = 0 + (-5\vec k) \times (-40\vec j) \\ v_c = -200\vec i\:  mm/svc​=vb​+wp​×rc/b​=0+(−5k)×(−40j​)vc​=−200imm/s
Now we can find an equation wa! 
﻿
wa=vcr1w_a = \frac {v_c}{r_1} wa​=r1​vc​​
﻿
Let's find Wa using the calculation tool.
Inputs﻿﻿vc
:-200​
﻿
Output﻿﻿wa
:-1.66666667​
﻿
﻿﻿r1
:120mm​
﻿
﻿

Final remark: Our final angular velocities for planet gear P and shaft are -5 rad/s and -1.67 rad/s respectively. 
﻿
Source: [4]
﻿
Continuous Loop Pulley SystemsA continuous loop pulley system is a type of pulley system that uses a continuous loop of rope to transmit power. This type of pulley system is commonly used in conveyor systems and is known for its ability to transmit power over long distances with minimal losses.
﻿
Continuous loop Pulley System
Worked Example: Continuous Pulley SystemA conveyor belt carrying aggregate is shown below. A motor turns the top roller at a constant speed, and the remaining rollers are allowed to spin freely. The belt is inclined at angle θ. To keep the belt in tension a weight of mass m is suspended from the belt, as shown. 
Find the point of maximum tension in the belt. You don't have to calculate it, just find the location and give a reason for it. 
﻿
Continuous Pulley System diagram
First Step: Make 6 imaginary cuts on the belt in the locations shown below
﻿
Continuous Pulley System diagram
Each T variable (T1, T2, T3, T4, T5, T6) represents the belt tensions.
﻿
Second Step: Perform analysis and reasoning 
﻿
First observation:
﻿
T1>T6T_1> T_6T1​>T6​
The tension at 1 supports the tension at 6 while supporting the weight between 1 and 6.

Second observation: 
﻿
T2>T3T_2 > T_3T2​>T3​
The tension at 2 has to support the tension at 3 while supporting the belt weight hanging between 2 and 3. 
﻿
Third observation: 
﻿
T3>T4T_3 > T_4T3​>T4​
T3 is higher due to the fact the length between the weight and 3 is longer than the length between the weight and 4.
﻿
Fourth observation:
﻿
T4>T5T_4 > T_5T4​>T5​
The tension at 4 has to support tension at 5 while supporting the belt weight between 4 and 5. 

Fifth observation:
﻿
T5=T6 (approx)T_5 = T_6 \: (approx)T5​=T6​(approx)
T5 and T6 are roughly equal due to their spatial locations and the fact that the roller spins freely.
﻿
Third Step: Perform analysis on the effect of the motor
﻿
The top roller is said to turn with constant rotational speed in a counterclockwise (CCW) direction, meaning a CCW torque due to the motor, is turning it in this direction. 

T1 exerts a CW torque on the top roller, and T2 exerts a CCW torque on the top roller. 
﻿
The roller is in rotational equilibrium as it is turning at a constant rotation speed. 
﻿
If we define CCW as positive and CW as negative we come up with the following equation:

﻿
MT+rT2−rT1=0MT=motor torquer=radius of top rollerMT + rT_2 - rT_1 = 0 \\ MT = motor \: torque\\ r = radius \: of \: top \: rollerMT+rT2​−rT1​=0MT=motortorquer=radiusoftoproller
We can produce the following equation:
﻿
T1=T2+(MT)/rT_1 = T_2 + (MT)/rT1​=T2​+(MT)/r
﻿
Final remark:
As T1 > T2, and using the equation above, it can therefore be concluded that the maximum tension in the belt is T1.
ConclusionPulleys serve as an essential component of today's mechanical systems, from construction to mechanical engineering to automotive design. Knowledge about pulleys and their different types is an essential piece of information for any aspiring engineer or technician in these fields. 
CalcTreeCalcTree, the app you're reading this one is a calculation management platform. You can sign-up and build hosted, shareable web apps (complete with an API and a web publishing module) with tools like Python and Spreadsheets. Learn more here!
References[1] https://www.real-world-physics-problems.com/pulley-problems.html#hint_for_problem_9
[2] Problem 16-77, Engineering Dynamics, 14th Edition, Russel C. Hibbeler
[3] https://www.youtube.com/watch?v=9Ar-nnI4jRk
﻿
﻿
﻿