In a hydraulic lift, the radii of the pistons are 2.65 cm and 9.2 cm. A car weighing W = 10.8 kN is to be lifted by the force of the large piston.
(a) What force Fa must be applied to the small piston? _____N
(b) When the small piston is pushed in by 8.0 cm, how far is the car lifted?____ mm
(c) Find the mechanical advantage of the lift, which is the ratio W / Fa.
(a) What force Fa must be applied to the small piston? _____N
(b) When the small piston is pushed in by 8.0 cm, how far is the car lifted?____ mm
(c) Find the mechanical advantage of the lift, which is the ratio W / Fa.
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internal pressure of hydraulic lift is constant
Fa / pi* 2.65^2 = 10800 / pi 9.2^2
Fa = 896N
answer
Volume traveled by smaller piston
= pi 2.65^2 * 8
= Volume traveled by bigger piston
= pi * 9.2^2 *h
pi 2.65^2 * 8 = pi * 9.2^2 *h
h = 0.664cm
= 6.64 mm
answer
mechanical advantage of the lift
= 10.8 / 0.896
= 12.05
answer
Fa / pi* 2.65^2 = 10800 / pi 9.2^2
Fa = 896N
answer
Volume traveled by smaller piston
= pi 2.65^2 * 8
= Volume traveled by bigger piston
= pi * 9.2^2 *h
pi 2.65^2 * 8 = pi * 9.2^2 *h
h = 0.664cm
= 6.64 mm
answer
mechanical advantage of the lift
= 10.8 / 0.896
= 12.05
answer