Register to reply

Force necessary to support mass on Hydraulic Jack?

by jumpingjack90
Tags: force, hydraulic, jack, mass, support
Share this thread:
jumpingjack90
#1
Aug17-09, 04:38 PM
P: 14
1. The problem statement, all variables and given/known data
Piston 1 in the figure has a diameter of 1.31 cm. distance of 2 in
Piston 2 has a diameter of 7.91 cm. distance of 10 in
In the absence of friction, determine the force F, necessary to support an object with a mass of 904 kg placed on piston 2. (Neglect the height difference between the bottom of the two pistons, and assume that the pistons are massless).

2. Relevant equations
F1*d1 = Fp*d2, where d1 is the moment arm of F1, and d2 is
the moment arm of Fp.

3. The attempt at a solution
Fp = (F1*d1)/d2
(904)(9.81)/(pi)(0.003955)^2= F/(pi)(0.00655)^2
F=242.98 N which was incorrect.
anyone have an idea of what I should do and what equation to use? Thanks!
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
Redbelly98
#2
Aug17-09, 07:32 PM
Mentor
Redbelly98's Avatar
P: 12,070
I think we'll need to see the figure before we can help you out.
jumpingjack90
#3
Aug17-09, 09:32 PM
P: 14
http://img6.imageshack.us/img6/7849/...56447261f7.gif

this is the link for the picture.

djeitnstine
#4
Aug17-09, 10:20 PM
PF Gold
djeitnstine's Avatar
P: 619
Force necessary to support mass on Hydraulic Jack?

Neglecting the height difference means the pressure should be the same, thus [tex]F_1 A_1 = F_2 A_2 [/tex]

This is would be the force exerted on the small piston.

You should know how to find the force needed by the hand from this.
Redbelly98
#5
Aug17-09, 11:33 PM
Mentor
Redbelly98's Avatar
P: 12,070
Quote Quote by jumpingjack90 View Post
2. Relevant equations
F1*d1 = Fp*d2, where d1 is the moment arm of F1, and d2 is
the moment arm of Fp.

3. The attempt at a solution
Fp = (F1*d1)/d2
(904)(9.81)/(pi)(0.003955)^2= F/(pi)(0.00655)^2
F=242.98 N which was incorrect.
anyone have an idea of what I should do and what equation to use? Thanks!
You have correctly found F1, the force on the smaller piston. However, they are asking for Fp (labelled simply F in the figure), the force exerted by the hand.

You have the equation to find Fp given F1.
jumpingjack90
#6
Aug18-09, 01:38 PM
P: 14
ok. I used the equation given and solved for Fp. I had (242.98)(pi*(0.03955)^2))/(pi)*(0.00655)^2=Fp=8858.92 N, which is also incorrect.
what am I doing wrong?
jumpingjack90
#7
Aug18-09, 03:23 PM
P: 14
nvm. I solved it! thanks everyone for your input!
george88b
#8
Mar1-11, 07:37 PM
P: 8
Quote Quote by jumpingjack90 View Post
nvm. I solved it! thanks everyone for your input!
how u solved it my friend?


Register to reply

Related Discussions
Looking for a Bottle Jack for a Hydraulic Car Jack Mechanical Engineering 27
Help me for calculate a Hydraulic force Mechanical Engineering 3
Quick hydraulic jack question Introductory Physics Homework 3
Hydraulic Car jack Mechanical Engineering 5
Building a hydraulic car jack Engineering, Comp Sci, & Technology Homework 0