Static equilibrium mass problem

[brentwoodbc]

Homework Statement

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?

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.

 

[tiny-tim]

Welcome to PF!

Hi brentwoodbc! Welcome to PF!

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 )

 

[brentwoodbc]

Hi brentwoodbc! Welcome to PF!


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 )

thanks, you I am 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]

… we were not given any example where the rod had a mass.

ok … the rod has a weight …

where does that weight act?

 

[brentwoodbc]

ok … the rod has a weight …

where does that weight act?

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]

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

 

[brentwoodbc]

so what's the equation going to look like?

F1+F2+F3=0?

Im lost on this question.

 

[tiny-tim]

No, torque₁ + torque₂ + torque₃ = 0

 

[brentwoodbc]

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]

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?

 

[brentwoodbc]

curiously, that seems to be the right answer …

but what was the (.45 -.25) supposed to be?

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)

 

[brentwoodbc]

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]

ah … got it!

that's fine then

 

[brentwoodbc]

ah … got it!

that's fine then

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.

 

[tiny-tim]

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.

oh, I missed the minus in your

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

that does make it .5

 

[brentwoodbc]

oh, I missed the minus in yourthat does make it .5

pfft. I'm so smart lol

thanks again though. Being spring break I can't 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.