Calcule as forças de avanço e recuo dos cilindros abaixo:

Para cilindros pneumáticos/hidráulicos, usamos:

• Área do êmbolo: $A_p=\dfrac{\pi D^2}{4}$
• Área da haste: $A_h=\dfrac{\pi d^2}{4}$
• Área efetiva no recuo (lado anular): $A_r=A_p-A_h$
• Força de avanço: $F_{av}=P\cdot A_p$
• Força de recuo: $F_{rec}=P\cdot A_r$

Onde $D$ é o diâmetro do êmbolo e $d$ o diâmetro da haste.

Conversões de pressão usadas:

• $1\,\text{bar}=10^5\,\text{Pa}$
• $1\,\text{psi}=6894{,}757\,\text{Pa}$
• $1\,\text{kPa}=10^3\,\text{Pa}$

A seguir, em cada item calculo $A_p$, $A_h$, $A_r$ e aplico $F=P\cdot A$ (sempre em SI: $P$ em Pa e $A$ em m², resultando $F$ em N).

1. $D=10\,\text{mm}=0{,}010\,\text{m}$, $d=4\,\text{mm}=0{,}004\,\text{m}$, $P=2\,\text{bar}=2\cdot10^5\,\text{Pa}$

• $A_p=\frac{\pi(0{,}010)^2}{4}=7{,}85398\times10^{-5}$
• $A_h=\frac{\pi(0{,}004)^2}{4}=1{,}25664\times10^{-5}$
• $A_r=6{,}59734\times10^{-5}$
• $F_{av}=2\cdot10^5\cdot A_p=15{,}71\,\text{N}$
• $F_{rec}=2\cdot10^5\cdot A_r=13{,}20\,\text{N}$

2. $D=20\,\text{mm}=0{,}020\,\text{m}$, $d=8\,\text{mm}=0{,}008\,\text{m}$, $P=58{,}0151\,\text{psi}$

• $P=58{,}0151\cdot6894{,}757=399999{,}96\,\text{Pa}\approx4{,}0\times10^5\,\text{Pa}$
• $A_p=\frac{\pi(0{,}020)^2}{4}=3{,}14159\times10^{-4}$
• $A_h=\frac{\pi(0{,}008)^2}{4}=5{,}02655\times10^{-5}$
• $A_r=2{,}63894\times10^{-4}$
• $F_{av}=P\cdot A_p=125{,}66\,\text{N}$
• $F_{rec}=P\cdot A_r=105{,}55\,\text{N}$

3. Mesmo $D$ e $d$ do item (1), mas $P=6\,\text{bar}=6\cdot10^5\,\text{Pa}$

• $F_{av}=6\cdot10^5\cdot7{,}85398\times10^{-5}=47{,}12\,\text{N}$
• $F_{rec}=6\cdot10^5\cdot6{,}59734\times10^{-5}=39{,}60\,\text{N}$

4. $D=3{,}2\,\text{cm}=0{,}032\,\text{m}$, $d=1{,}2\,\text{cm}=0{,}012\,\text{m}$, $P=87{,}0226\,\text{psi}$

• $P=87{,}0226\cdot6894{,}757=599999{,}99\,\text{Pa}\approx6{,}0\times10^5\,\text{Pa}$
• $A_p=\frac{\pi(0{,}032)^2}{4}=8{,}04248\times10^{-4}$
• $A_h=\frac{\pi(0{,}012)^2}{4}=1{,}13097\times10^{-4}$
• $A_r=6{,}91150\times10^{-4}$
• $F_{av}=P\cdot A_p=498{,}23\,\text{N}$
• $F_{rec}=P\cdot A_r=428{,}17\,\text{N}$

5. $D=6{,}3\,\text{cm}=0{,}063\,\text{m}$, $d=2{,}0\,\text{cm}=0{,}020\,\text{m}$, $P=6\,\text{bar}=6\cdot10^5\,\text{Pa}$

• $A_p=\frac{\pi(0{,}063)^2}{4}=3{,}11725\times10^{-3}$
• $A_h=\frac{\pi(0{,}020)^2}{4}=3{,}14159\times10^{-4}$
• $A_r=2{,}80309\times10^{-3}$
• $F_{av}=6\cdot10^5\cdot A_p=1870{,}35\,\text{N}$
• $F_{rec}=6\cdot10^5\cdot A_r=1681{,}86\,\text{N}$

6. $D=160\,\text{mm}=0{,}160\,\text{m}$, $d=40\,\text{mm}=0{,}040\,\text{m}$, $P=10\,\text{bar}=10^6\,\text{Pa}$

• $A_p=\frac{\pi(0{,}160)^2}{4}=2{,}01062\times10^{-2}$
• $A_h=\frac{\pi(0{,}040)^2}{4}=1{,}25664\times10^{-3}$
• $A_r=1{,}88496\times10^{-2}$
• $F_{av}=10^6\cdot A_p=20106{,}19\,\text{N}$
• $F_{rec}=10^6\cdot A_r=18849{,}56\,\text{N}$

7. $D=0{,}32\,\text{m}$, $d=0{,}063\,\text{m}$, $P=1000\,\text{kPa}=10^6\,\text{Pa}$

• $A_p=\frac{\pi(0{,}32)^2}{4}=8{,}04248\times10^{-2}$
• $A_h=\frac{\pi(0{,}063)^2}{4}=3{,}11725\times10^{-3}$
• $A_r=7{,}73075\times10^{-2}$
• $F_{av}=10^6\cdot A_p=80424{,}77\,\text{N}$
• $F_{rec}=10^6\cdot A_r=77307{,}88\,\text{N}$

Como não há alternativas de múltipla escolha, o resultado final são as forças de avanço e recuo em newtons para cada exercício.

Alternativa correta: (sem alternativas).