B Relating displacements in a pulley system

AI Thread Summary
In the discussion about a pulley system, a force \( F \) pulls a rope down by a distance \( dx_1 \), causing a mass \( M \) to rise by \( dx_2 \). The energy conservation principle is applied, leading to the equation \( Tdx_1 = 2Tdx_2 - Mg dx_2 \). This is explained by noting that the energy input from the force \( F \) translates to tension \( T \) in the rope. The derived relationship simplifies to \( dx_1 = (2 - \frac{Mg}{T}) dx_2 \), but there is confusion about whether this can be simplified to \( dx_1 = 2 dx_2 \) based on rope conservation principles. The discussion seeks clarification on the conceptual understanding of these relationships.
burian
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> The set up: At left end, the rope is pulled down with a distance $dx_1$ by a force of constant magnitude $F$, the mass of $M$ is wrapping around by rope on the right and moves up by a distance $dx_2$ due to this. Problem: Find relate the two displacements.

I thought of applying energy conservation, we put in energy $Tdx_1$ into the system and we add up the energy induced on the rest of the system. Noting that $F=T$,

$$Tdx_1= 2Tdx_2 - Mg dx_2$$

**Explanation for left side** : We input an energy of $F \cdot dx_1$ when we pull the rope by$dx_1$ with force $F$, since $F=T$, the energy is just $T dx_1$

**Explanation for right side :** The mass is pulled up by $dx_2$, this goes into kinetic energy of the body, this kinetic energy can be written using the work energy theorem as the external forces dotted with $dx_2$, $(2T-mg) \hat{j} \cdot (dx_2 \hat{j})= (2T- mg) dx_2$

This simplifies to:

$$ dx_1 = (2- \frac{Mg}{T} ) dx_2 \tag{1}$$
But, if we go by the differential conservation of rope, we find that $dx_1 = 2 dx_2$ is it possible to simplfy eqtn (1) into this, or have I done something conceptually wrong?
 

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Not sure why inline latex is not working..
 
You need to use double $ sign.
$$F=T$$
 
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