File:PulleySol1.jpeg
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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]
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current | 17:10, 21 August 2017 | 2,321 × 2,962 (387 KB) | Raxu (talk | contribs) | 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... |
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