There are two commonly used elements, levers and pulleys, that are used to amplify forces or displacements. There is a direct proportionate tradeoff between increase in force and decrease in displacement (or viceversa), but power is preserved (i.e., the power on the input side of one of these devices is equal to the power out). The behavior and modeling of these elements is the focus of this section.
Contents
Consider a massless lever as shown below.
Lever  Geometric Relationships 
L is the ratio of lever arms. 
This states that the ratio of the displacement of x_{1} relative to x_{2} is simply the ratio, L, of the two lever arms ℓ_{1} and ℓ_{2}.
We can find a relationship between forces and displacements by summing the torques around the fulcrum. We also make the small angle approximation so we can assume the force is perpendicular to the lever, and that the motion at the ends of the lever is purely translational in the "x" direction. (Without this approximation we must multiply the magnitude of the force by cos(θ) to get the force perpendicular to the lever.) To do this we add counterclockwise (ccw) torques and subtract clockwise (cw) torques and sum to zero.
This states that the ratio of the force f_{1} relative to f_{2} is simply the inverse of the ratio of the two lever arms ℓ_{1} and ℓ_{2}.
Note that if we combine the force relationship with the velocity relationship, we get:
You will see on the next web page that the product of velocity and force is power. So the last equation simply states that power is preserved; it is the same on both sides of the lever. The lever allows one to trade force for velocity (i.e., if the lever arms are unequal, one side of the lever will experience higher force but lower velocity than the other side).
Develop a mathematical model for the system shown in terms of f_{a} and x_{1}.
We can draw a freebody diagram with horizontal forces and sum the torques (note: we don't care about the force from the fulcrum since its moment arm is zero because we are summing about the fulcrum):
We can rewrite the relationship from the previous example in terms of the position at ℓ_{1} (associated with displacement x_{1}) at which the force acts. This requires some simple manipulations of the equation of motion from the example.
This means that we can create an equivalent system by moving the spring from the location at ℓ_{2} (associated with displacement x_{2}) to the location at ℓ_{1} (associated with displacement x_{1}) by multiplying the spring constant by N^{2} where N=ℓ_{2}/ℓ_{1}, (i.e, N is the ratio of the original moment arm to the moment arm at the new position) In other words, the two systems below are equivalent.
Original System  Equivalent system 
Likewise, we could write the equations in terms of x_{2}, and develop another equivalent system.
Rewrite Equations in terms of x_{2} 
Original System  Equivalent System 
To generalize: a spring, mass or friction element can be translated along a lever from a position with a distance from the fulcrum of ℓ_{1} to a distance of ℓ_{2} by multiplying by N^{2}, where N=ℓ_{1}/ℓ_{2},. A force can be moved from a distance of ℓ_{1} to ℓ_{2} by multiplying by N. An example will clarify.
Develop an equivalent mechanical model at the position x_{2}.
We can effectively move the spring from x_{3} to x_{2} by multiplying by (ℓ_{3}/ℓ_{2})^{2}. We can effectively move the force from x_{1} to x_{2} by multiplying by (ℓ_{1}/ℓ_{2}). This yields the equivalent system below (along with free body diagram and mathematical model).
Equivalent System  Equations of Motion 
Check: As a quick check, let's make free body diagram of original system, sum torques, and simplify. The results must be the same as those above if this technique is correct.
Free Body Diagram  Equations of Motion 
(note negative sign in relationship between x_{2} and x_{3}) 
A spring, mass or friction element can be translated along a lever from a position with a distance from the fulcrum of ℓ_{1} to a distance of ℓ_{2} by multiplying by N^{2}, where N=ℓ_{1}/ℓ_{2},. A force can be moved from a distance of ℓ_{1} to ℓ_{2} by multiplying by N.
Massless pulleys act in very much the same way as the massless lever discussed previously. Consider the system shown below. The pulley has two radii, r_{1} and r_{2}.
If the pulley moves through an angle θ the distance x_{1} is given by r_{1}·θ, and the distance x_{2} is given by r_{2}·θ. Note: The relationship(s) between position and angles are always necessary to solve these systems; if it looks like you don't have enough equations for a solution, make sure you are using those relationships.
Recall that the arc length along the circumference of a circle is equal to r·θ where r is the radius of the circle and theta is the angle around the circle. If a rope is wrapped around the circle and the circle rotates through an angle θ, the end of the rope will move by an amount r·θ. It is important to be careful about the sign of the relationship.
If the rope comes off the top, and the distance is defined as positive to the left, then x_{1}=r·θ. If rope comes off the top and the distance is defined as positive to the right, then x_{2}=r·θ. See below.
It now becomes apparent that a pulley behaves in the same manner as a lever because the relative displacements are determined by the ratio of the radii (for the lever it was the ratio of the lever arms). However, in this case there is no need to invoke the small angle argument since all motion is purely translational.
A spring, mass or friction element connected to a pulley through a radius r_{1} can be translated to the another pulley with radius r_{2} by multiplying by N^{2}, where N=r_{1}/r_{2}. A force can be moved from r_{1} to r_{2} by multiplying by N.
Let's demonstrate these ideas with an example to be solved twice. The first solution will sum torques on the pulley, the second time will use effective elements.
Develop a mathematical model in terms of the position x_{2}. Take the equilibrium position of x_{1} and x_{2} to be 0 when f_{a}=0.
Since the equilibrium position is defined to be zero we need not consider gravity in our model (reference). Let's draw free body diagrams (one for x_{1}, x_{2} and θ)
x_{1}  θ  x_{2}  
Free Body Diagram 
T_{1} is the tension in the upper rope. 
T_{2} is the tension in the lower rope. 

Equations of Motion 
To solve the equations, we also need to use the relationships between x_{1}, x_{2} and θ.
To solve we will use the torque relationship to solve for T_{1} in terms of T_{2}, and use the geometric relationship to solve for x_{1} in terms of x_{2}.
Now we can substitute these into the free body equation at x_{1}, and solve for T_{2} in terms of x_{2}
Substitute this in the free body equation at x2, and we are done
Repeat the previous example using effective elements.
We can translate the mass m_{1}, friction b, and spring k from x_{1} to x_{2} by multiplying by N^{2}, where N=r_{1}/r_{2}.
x_{2}  
Free Body Diagram 

Equations of Motion 
The two methods yield equivalent results.
Obviously the second method is computationally simpler; but conceptually more difficult. You should check your results by summing torques until you feel comfortable applying the concept of "effective" elements (either with either levers or pulleys).
© Copyright 20052013 Erik Cheever This page may be freely used for educational purposes.
Erik Cheever Department of Engineering Swarthmore College