Exploring the Physics Behind Fixed and Movable Pulleys

[mathmari]

Hey!

View attachment 4338

Fixed pulley:

$$F_G \cdot r=F_z \cdot r \Rightarrow F_G=F_z$$

Only force transfer, no changement of force

Movable pulley:

Like one-sided lever, current center of rotation is at the circumference of the pulley:

$$F_G \cdot r=F_z \cdot 2r \Rightarrow F_z=\frac{F_G}{2}$$

Could you explain me why these formulas stand?? (Wondering)

 

[Fantini]

You assume the systems are in equilibrium. Therefore, the sum of the total torques of each system must be zero.

In the fixed pulley you measure torques with respect to the center of the pulley. In this situation, the torque due to the man is $\tau_m = F_z \cdot r$ in the counterclockwise orientation. The torque due to the box is $\tau_b = F_G \cdot r$, in the clockwise orientation. The sum of these torques is $$\tau_m - \tau_b = F_z \cdot r - F_G \cdot r = 0.$$The negative sign is due to the convention that clockwise orientation gets a negative sign, just as in trigonometry. You obtain the desired result.

There is another way to view this. The thread is massless, and this means the force is fully transmitted to the mass. Therefore, the box has two forces acting on it: the weight and the pulling force. Both forces act vertically, with the weight pointing down and the pulling force pointing up. Since it is in equilibrium we have $F_z - F_G = 0$.

The same idea is applied in the second case, but the point of reference is not the center of the second pulley. Instead you take the point of reference on the right side of the pulley. The distance to the pulling force $F_z$ is $2r$, now acting in a clockwise orientation, and the distance to the weight $F_G$ is $r$, acting in a counterclockwise orientation. Again we assume equilibrium and therefore $$-F_z \cdot 2r + F_G \cdot r = 0.$$