Posts Tagged ‘fixed pulley’

More Pulleys Increase Mechanical Advantage

Thursday, October 27th, 2016

    Last time we saw how compound pulleys within a dynamic lifting scenario result in increased mechanical advantage to the lifter, mechanical advantage being an engineering  phenomenon that makes lifting weights easier.   Today we’ll see how the mechanical advantage increases when more fixed and movable pulleys are added to the compound pulley arrangement we’ve been working with.

More Pulleys Increase Mechanical Advantage

More Pulleys Increase Mechanical Advantage


    The image shows a more complex compound pulley than the one we previously worked with.   To determine the mechanical advantage of this pulley, we need to determine the force, F5, Mr. Toga exerts to hold up the urn.

    The urn is directly supported by four equally spaced rope sections with tension forces F1, F2, F3, and F4.   The weight of the urn, W, is distributed equally along the rope, and each section bears one quarter of the load.   Mathematically this is represented by,

F1 = F2 = F3 = F= W ÷ 4

    If the urn’s weight wasn’t distributed equally, the bar directly above it would tilt.   This tilting would continue until equilibrium was eventually established, at which point all rope sections would equally support the urn’s weight.

    Because the urn’s weight is equally distributed along a single rope that’s threaded through the entire pulley arrangement, the rope rule, as I call it, applies.   The rule posits that if we know the tension in one section of rope, we know the tension in all rope sections, including the one Mr. Toga is holding onto.   Therefore,

F1 = F2 = F3 = F4 = F5 = W ÷ 4

    Stated another way, the force, F5 , Mr. Toga must exert to keep the urn suspended is equal to the weight force supported by each section of rope, or one quarter the total weight of the urn, represented by,

F5 = W ÷ 4

    If the urn weighs 40 pounds, Mr. Toga need only exert 10 pounds of bicep force to keep it suspended, and today’s compound pulley provides him with a mechanical advantage, MA, of,

MA = W ÷  F5

MA = W ÷  (W ÷ 4)

MA = 4

    It’s clear that adding the two extra pulleys results in a greater benefit to the man doing the lifting, decreasing his former weight bearing load by 50%.   If we added even more pulleys, we’d continue to increase his mechanical advantage, and he’d be able to lift far heavier loads with a minimal of effort.   Is there any end to this mechanical advantage?  No, but there are undesirable tradeoffs.   We’ll see that next time.

 Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



The Math Behind a Static Compound Pulley

Friday, September 9th, 2016

    Last time we introduced the compound pulley and saw how it improved upon a simple pulley, both of which I’ve engaged in my work as an engineering expert.  Today we’ll examine the math behind the compound pulley.   We’ll begin with a static representation and follow up with an active one in our next blog.

    The compound pulley illustrated below contains three rope sections with three representative tension forces, F1, F2, and F3.   Together, these three forces work to offset the weight, W, of a suspended urn weighing 40 lbs.   Weight itself is a downward pulling force due to the effects of gravity.

    To determine how our pulley scenario affects the man holding his section of rope and exerting force F3, we must first calculate the tension forces F1 and F2.   To do so, we’ll use a free body diagram, shown in the green box, to display the forces’ relationship to one another.

A Compound Pulley

The Math Behind a Static Compound Pulley

    The free body diagram only takes into consideration the forces inside the green box, namely F1, F2, and W.

    For the urn to remain suspended stationary in space, we know that F1 and F2 are each equal to one half the urn’s weight, because they’re spaced equidistant from the pulley’s axle, which directly supports the weight of the urn.  Mathematically this looks like,

F1 = F2 = W ÷ 2

    Because we know F1 and F2, we also know the value of F3, thanks to an engineering rule concerning pulleys.  That is, when a single rope is used to support an object with pulleys, the tension force in each section of rope must be equal along the entire length of the rope, which means F1 = F2 = F3.    This rule holds true whether the rope is threaded through one simple pulley or a complex array of fixed and moveable simple pulleys within a compound pulley.   If it wasn’t true, then unequal tension along the rope sections would result in some sections being taut and others limp, which would result in a situation which would not make lifting the urn any easier and thereby defeat the purpose of using pulleys.

    If the urn’s weight, W, is 40 pounds, then according to the aforementioned engineering rule,

F1 = F2 = F3= W ÷ 2

F1 = F2 = F3 =  (40 pounds) ÷ 2 = 20 pounds

    Mr. Toga needs to exert a mere 20 pounds of personal effort to keep the immobile urn suspended above the ground.   It’s the same effort he exerted when using the improved simple pulley in a previous blog, but this time he can do it from the comfort and safety of standing on the ground.

    Next time we’ll examine the math and mechanics behind an active compound pulley and see how movement affects F1 , F2 , and F3.


Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog