Factorial Rigging: Definition, Formulas and Exercises
The factorial probe is a simple machine consisting of an arrangement of pulleys with a force multiplier effect. In this way, a load can be lifted by applying only the equivalent of a fraction of the weight to the free end of the cable.
It consists of two sets of pulleys: one fixed to a support and another that exerts the resulting force on the load. The pulleys are mounted on a generally metallic frame that holds them.
Formulas for factorial rigging
Case 1: A mobile pulley and a fixed pulley
To understand why this arrangement multiplies the exerted force, we’ll start with the simplest case, which consists of a fixed pulley and a movable pulley.
Suppose we are in an equilibrium situation. Consider the forces acting on pulley B. The shaft of pulley B supports a total weight P that is directed downward. If this were the only force on pulley B, it would fall, but we know that the cable passing through this pulley also exerts two forces, which are T1 and T2 directed upwards.
For the translation balance to exist, the two upward forces must be equal to the weight that supports the pulley B axis.
T1 + T2 = P
But since pulley B is also in rotational equilibrium, so T1 = T2. Forces T1 and T2 come from the tension applied to the cable, called T.
So T1 = T2 = T. Substituting it in the previous equation, it is:
T + T = P
2T = P
Which indicates that the tension applied to the cable is only half the weight:
T = P / 2
For example, if the load were 100 kg, it would suffice to apply a force of 50 kg at the free end of the rope to increase the load at a constant speed.
Case 2: two movable pulleys and two fixed pulleys
Let’s consider the stresses and forces acting on a set consisting of two arrangements of supports A and B with two pulleys each.
Figure 3. Forces on a platform with 2 fixed pulleys and 2 mobile pulleys.
Support B has the ability to move vertically, and the forces acting on it are:
– The weight P of the load, pointing vertically downwards.
– Two tensions on the large pulley and two tensions on the small pulley. In total, four tensions, all pointing upwards.
For translational equilibrium, it is necessary that the forces that point vertically upward are equal in value to the load that points downward. That is, it must be fulfilled:
T + T + T + T = P
That is, 4 T = P
As a result, the applied force T at the free end of the rope is only a quarter of the weight due to the load it wishes to lift., T = P / 4.
With this value for the voltage T, the load can be kept static or raised at a constant speed. If a voltage greater than this value is applied, the load will accelerate upward, a necessary condition to remove it from rest.
General case: n movable pulleys n fixed pulleys
As seen in the previous cases, for each pulley of the movable assembly, there are some upward forces exerted by the cable passing through the pulley. But this force cannot be anything other than the tension applied to the cable at the free end.
Therefore, for each pulley in the movable assembly, there will be an upward vertical force worth 2T. However, as there are n pulleys in the movable assembly, it is necessary that the total force pointed vertically upwards is:
2 n T
In order to have vertical balance, it is necessary that:
2 n T = P
therefore, the force applied at the free end is:
T = P / (2 n)
In this case, it can be said that the exerted force T is multiplied 2 n times in the load.
For example, if we had a factorial platform of 3 fixed pulleys and 3 movable pulleys, the number n would be equal to 3. On the other hand, if the load were P = 120 kg, the force applied at the free end would be T = 120 kg / ( 2 * 3) = 20 kg.
Consider a factorial platform composed of two fixed pulleys and two mobile pulleys. The maximum tension that the cable can withstand is 60 kg. Determine the maximum load that can be placed.
When the load is at rest or moving at a constant speed, its weight P is related to the voltage T applied to the cable through the following relationship:
P = 2 n T
As it is a platform with two movable pulleys and two fixed pulleys, then n = 2.
The maximum load that can be placed is obtained when T has the maximum possible value, which in this case is 60 kg.
Maximum load = 2 * 2 * 60 kg = 240 kg
Find the relationship between cable tension and load weight, in a bipolar factorial platform, where the load accelerates with acceleration a.
The difference in this example from what has been seen so far is that the dynamics of the system must be considered. So, we propose Newton’s second law to find the requested relationship.
Y1 is the position of the lowest pulley axis.
We apply Newton’s second law to determine the acceleration a1 of the moving part of the platform:
-4 T + Mg = M a1
Since the load weight is P = Mg, where g is the acceleration due to gravity, the above relationship can be written:
-4T + P = P (a1 / g)
If we wanted to determine the voltage applied to the cable when a certain load of weight P accelerates with acceleration a1, the previous relationship would be like this:
T = P (1 – a1 / g) / 4
Note that if the system were at rest or moving at constant speed, then a1 = 0, and we would retrieve the same expression we got in case 2.
In this example, the same platform as in exercise 1 is used, with the same rope that supports a maximum of 60 kg of tension. A certain load increases, accelerating it from rest to 1 m / s in 0.5 s, using the maximum cable tension. Find the maximum load weight.
We will use the expressions obtained in exercise 2 and the frame of reference in Figure 4, in which the positive direction is vertical in the vertical.
The load acceleration is a1 = (-1 m / s – 0 m / s) / 0.5 s = -2 m / s ^ 2.
The load weight in kilograms-force is given by
P = 4 T / (1 – a1 / g)
P = 4 * 60 kg / (1 + 2 / 9.8) = 199.3 kg
This is the maximum possible weight of the load without breaking the rope. Note that the value obtained is smaller than that obtained in example 1, in which the load was assumed with zero acceleration, that is, at rest or with constant velocity.