The areas of the pistons in a hydraulic machine are $5 {c m}^{2}$ and $625 {c m}^{2}$ . What force on the smaller piston will support a load of $1250 N$ on the larger piston?

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Solution

$F_{1} = 1250 N A_{1} = 625 {c m}^{2}$
$F_{2} = x n A_{2} = 5 {c m}^{2}$
Since pressure is constant.
$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \Rightarrow \frac{1250}{625} = \frac{x}{5} \Rightarrow x = 10 N$

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The areas of the pistons in a hydraulic machine are 5cm2 and 625cm2. What force on the smaller piston will support a load of 1250N on the larger piston?

F1=1250NA1=625cm2 F2=xnA2=5cm2Since pressure is constant.F1A1=F2A2⇒1250625=x5⇒x=10N

The areas of the pistons in a hydraulic machine are $5 {c m}^{2}$ and $625 {c m}^{2}$ . What force on the smaller piston will support a load of $1250 N$ on the larger piston?

$F_{1} = 1250 N A_{1} = 625 {c m}^{2}$
$F_{2} = x n A_{2} = 5 {c m}^{2}$
Since pressure is constant.
$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \Rightarrow \frac{1250}{625} = \frac{x}{5} \Rightarrow x = 10 N$