Answer :
Answer:
1. 88370.45 N/[tex]m^{2}[/tex]
2. 2500 N.
Explanation:
Pressure, P, is define as the force, F, per unit area, A, applied on/ by an object.
i.e P = [tex]\frac{F}{A}[/tex]
1. The area, A, of the piston of the hydraulic elevator can be determined by;
A = [tex]\pi[/tex][tex]r^{2}[/tex]
where r is the radius of the piston.
A = [tex]\frac{22}{7}[/tex] × [tex](0.3)^{2}[/tex]
= [tex]\frac{22}{7}[/tex] × 0.09
= 0.2829 [tex]m^{2}[/tex]
Pressure required to lift a truck of 2500 kg mass can be determined as;
P = [tex]\frac{F}{A}[/tex]
F = W = mg
= 2500 × 10
= 25 000 N
So that,
P = [tex]\frac{25000}{0.2829}[/tex]
= 88370.45 N/[tex]m^{2}[/tex]
The pressure required is 88370.45 N/[tex]m^{2}[/tex].
2. [tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
Area of small piston = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] × [tex](0.03)^{2}[/tex]
= 0.02829 [tex]m^{2}[/tex]
So that,
[tex]\frac{25000}{0.2829}[/tex] = [tex]\frac{F_{2} }{0.02829}[/tex]
[tex]F_{2}[/tex] = 2500 N
The force applied to the small piston is 2500 N.