The radii of the press plunger and the pump plunger are in the ratio $30 : 4$ . If an effort of $32 k g f$ acts on the pump plunger. Find the maximum effort the press plunger can overcome. ( in kgf)

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Solution

$\frac{r_{1}}{r_{2}} = \frac{30}{4} & F_{2} = 32 k g f$
to find $F_{1}$
Since from pascals law
$P_{1} = P_{2} \Rightarrow \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \Rightarrow F_{1} = F_{2} \times \frac{A_{1}}{A_{2}} = F_{2} \times {(\frac{r_{1}}{r_{2}})}^{2}$
$F_{1} = 32 \times \frac{15^{2}}{4} = 1800 k g f$

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The radii of the press plunger and the pump plunger are in the ratio 30:4. If an effort of 32kgf acts on the pump plunger. Find the maximum effort the press plunger can overcome. ( in kgf)

r1r2=304&F2=32kgf to find F1Since from pascals lawP1=P2⇒F1A1=F2A2⇒F1=F2×A1A2=F2×(r1r2)2 F1=32×1524=1800kgf

The radii of the press plunger and the pump plunger are in the ratio $30 : 4$ . If an effort of $32 k g f$ acts on the pump plunger. Find the maximum effort the press plunger can overcome. ( in kgf)