## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 9 - Problems - Page 360: 10

#### Answer

(a) A force of $625~N$ must be applied to the small piston. (b) The car is lifted 6.25 mm (c) The mechanical advantage is 16

#### Work Step by Step

(a) We can use Pascal's principle to find $F_a$: $\frac{F_a}{a} = \frac{F_A}{A}$ $\frac{F_a}{\pi~r_a^2} = \frac{F_A}{\pi~r_A^2}$ $F_a = \frac{r_a^2~F_A}{r_A^2}$ $F_a = \frac{(2.50~cm)^2~(10,000~N)}{(10.0~cm)^2}$ $F_a = 625~N$ A force of $625~N$ must be applied to the small piston. (b) The work done on the small piston is equal to the work done by the large piston on the car: $F_A~d_A = F_a~d_a$ $d_A = \frac{F_a~d_a}{F_A}$ $d_A = \frac{(625~N)(10.0~cm)}{10,000~N}$ $d_A = 0.625~cm = 6.25~mm$ The car is lifted 6.25 mm (c) We can find the mechanical advantage: $\frac{W}{F_a} = \frac{10000~N}{625~N} = 16$ The mechanical advantage is 16.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.