Static equilibrium mass problem

1. The problem statement, all variables and given/known data
A uniform 0.122 kg rod of 0.90 m length is used to suspend two masses as shown.
At what distance x should the 0.20 kg mass be placed to achieve static equilibrium?
4010113.gif



2. The attempt at a solution

F1=1.96N
F2=4.9N

torque = 4.9 x 0.25 = 1.225 Nm.cw

1.23 = 1.96r

r=0.63m


answer is actually 0.5

I think its because of the force of 1.96 (0.20kg) being to the right of the end of the rod (which has a mass)


could someone please help me.
 
Last edited:

tiny-tim

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Welcome to PF!

Hi brentwoodbc! Welcome to PF! :smile:
F1=1.96N
F2=4.9N
What about F3 (for the rod)?

(And please don't multiply everything by 9.8 …

just call it g … all the g's wil cancel in the end, anyway :rolleyes:)
 
Re: Welcome to PF!

Hi brentwoodbc! Welcome to PF! :smile:


What about F3 (for the rod)?

(And please don't multiply everything by 9.8 …

just call it g … all the g's wil cancel in the end, anyway :rolleyes:)

thanks, ya Im trying to figure out how to factor in the force of the rod but we were not given any example where the rod had a mass.
 

tiny-tim

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Re: Equalibrium

ok … the rod has a weight …

where does that weight act? :smile:
on the centre of rotation. and the sum of all forces = zero. What direction is the 3rd force though? against the other two?
 

tiny-tim

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on the centre of rotation. and the sum of all forces = zero. What direction is the 3rd force though? against the other two?
(sorry, i'm not following you)

the weight of a rod (or indeed anything else) acts through its centre of mass
 
Re: Equalibrium

so whats the equation going to look like?

F1+F2+F3=0?

Im lost on this question.
 

tiny-tim

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No, torque1 + torque2 + torque3 = 0 :smile:
 
Re: Equalibrium

ok, but I keep getting -0.35

.5gr+.2gr+.122gr=0

.2gr=-(.122g(.45 -.25)+.5(.25))

cancel g

divide .2

r = 0.75?
 

tiny-tim

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ok, but I keep getting -0.35

.5gr+.2gr+.122gr=0

.2gr=-(.122g(.45 -.25)+.5(.25))

cancel g

divide .2

r = 0.75?
curiously, that seems to be the right answer …

but what was the (.45 -.25) supposed to be? :confused:
 
Re: Equalibrium

curiously, that seems to be the right answer …

but what was the (.45 -.25) supposed to be? :confused:
half of distance of beam is 0.45metres

so minus the 0.25metres is the distance of the centre of mass from the centre of rotation so (r for F3)
 
Re: Equalibrium

I noticed that the correct answer "0.5"is my answer "0.75" minus the 0.25 to the right of the centre of rotation... hmmm.
 

tiny-tim

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ah … got it!

that's fine then :smile:
 
Re: Equalibrium

ah … got it!

that's fine then :smile:
Im not 100% but I think that since the force with mass.5 is clockwise "in direction" it is negative. the other two are in a counter clockwise direction. So I made the 0.25metres negative and I got 0.5.

Seems to be fine. Thank you very much.:biggrin:
 
Re: Equalibrium

oh, I missed the minus in your


that does make it .5 :redface:
pfft. I'm so smart lol :cool:

thanks again though. Being spring break I cant get help and I have tests a couple days after I go back. And there are more question s in my homework similar to this so I should be cool now.
 

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