File:PulleySol1.jpeg

Let the upper pulley be pulley 2, and the lower pulley be pulley 1. [math] G [/math] is the weight of the load. Method 1: using forces. Since The two upwards tension forces are from the same string, they are equal. Let that force be [math]T[/math]. By drawing the free body diagram of pulley 1, we find that [math]T+T-G=0, T=\frac{G}{2}.[/math] Then, since [math]F_{in}[/math] is from the same string as the tension, [math]F_{in}=T=\frac{G}{2}, IMA = 2.[/math]

Method 2: Using distances. When the load moves upward by [math]d_{out}[/math], pulley 1 moves upward by [math]d_1=d_{out}[/math]. Therefore, the length of the middle and right segment of the string is each reduced by [math]d_1[/math]. Since the total length of the string stays the same, the length of left segment must increase by [math]d_{in}=2d_1=2d_{out}, IMA = 2.[/math]