In a hydraulic lift, the radii of the pistons are 2.93 cm and 10.1 cm. A car weighing W = 11.0 kN is to be lifted by the force of the large piston.
(a) What force Fa must be applied to the small piston? 925.7 N
(b) When the small piston is pushed in by 10.4 cm, how far is the car lifted? mm
b. h2 = h1•(A1/A2) = h1•(r/R) = 3.02 cm
(c) Find the mechanical advantage of the lift, which is the ratio W / Fa.
11.9
Sorry for reposting, but i don't know why part b is wrong the answer has to be in mm, so 3.02 cm would be 30.2 mm, but it giving me wrong; please help. Thank you.
part b is wrong because you did (A1/A2)=(r/R)
the ratio of radii should be squared...
Thank you!
To help you find the correct answer for part (b) of the question, let's go through the steps again.
In a hydraulic lift, the relationship between the heights of the liquid on the two sides is given by:
h2 = h1 * (A1 / A2)
where:
h2 = height of the lift on the large piston side
h1 = height of the lift on the small piston side
A1 = cross-sectional area of the small piston
A2 = cross-sectional area of the large piston
Given that the radius of the small piston is 2.93 cm and the radius of the large piston is 10.1 cm, we can calculate the cross-sectional areas:
A1 = π * (2.93 cm)^2
A2 = π * (10.1 cm)^2
Now, let's substitute these values into the equation and calculate h2:
h2 = h1 * (A1 / A2)
h2 = h1 * ((π * (2.93 cm)^2) / (π * (10.1 cm)^2))
h2 = h1 * (2.93^2 / 10.1^2)
Now, plug in the value of h1, which is given as 10.4 cm:
h2 = 10.4 cm * (2.93^2 / 10.1^2)
Calculating this expression will give you the correct answer for part (b), in millimeters.