Question

A hydraulic jack used to lift cars is called a three-ton jack.
The large piston is 22 mm in diameter, the small one 6.3 mm. Assume
that a force of 3 tons is 3.0 x 10^{4} N.

a. What force must be exerted on the small piston to lift the
3-ton weight?

b. Most jacks use a lever to reduce the force needed on the small
piston. If the resistance arm is 3.0 cm, how long is the effort arm
of an ideal lever to reduce the force to 100.0 N?

Answer #1

a. What force must be exerted on the small piston to lift the 3-ton weight?

F_{1} / A_{1} = F_{2} /
A_{2}

So F_{2} = F_{1}A_{2} / A_{1} =
F_{1}
(r_{2})^{2} /
(r_{1})^{2} = F_{1}( r_{2} /
r_{1})^{2}

F_{2} = (3.0 x 10^{4} N) [(6.3 mm /2)/(22mm
/2)]^{2} = 2.5 x 10^{3} N

b. Most jacks use a lever to reduce the force needed on the small piston. If the resistance arm is 3.0 cm, how long is the effort arm of an ideal lever to reduce the force to 100.0 N?

MA = F_{r} / F_{e} = 2.5 x 10^{3} N /
100.0 N = 25

and IMA = L_{e} / L_{r} with MA = IMA

L_{e} = (MA)L_{r} = 25(3.0 cm) = 75 cm

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