Question #21241

1 Answer
P dilip_k · Grace
Feb 13, 2016

The force applied at smaller piston = #20# #kg#

Explanation:

By Pascal's law of multiplication of thrust we have

#F_2 = F_1 " x " (A_2/A_1)#
where force exerted on larger piston (#F_2#) = 1600 #kg#
force exerted on smaller piston (#F_1#) = unknown
area of larger piston (#A_2#) = 800 #cm^2#
area of larger piston (#A_1#) = 10 #cm^2#

#F_1 = F_2 " x " (A_1/A_2)#
= #1600 " x "10/800#
= #20# #kg#

Related questions
Impact of this question
815 views around the world
You can reuse this answer
Creative Commons License