Static equilibrium mass problem

In summary, 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?
  • #1
brentwoodbc
62
0

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?
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:
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  • #2
Welcome to PF!

Hi brentwoodbc! Welcome to PF! :smile:
brentwoodbc said:
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:)
 
  • #3


tiny-tim said:
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, 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.
 
  • #4
brentwoodbc said:
… we were not given any example where the rod had a mass.

ok … the rod has a weight …

where does that weight act? :smile:
 
  • #5


tiny-tim said:
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?
 
  • #6
brentwoodbc said:
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
 
  • #7


so what's the equation going to look like?

F1+F2+F3=0?

Im lost on this question.
 
  • #8
No, torque1 + torque2 + torque3 = 0 :smile:
 
  • #9


ok, but I keep getting -0.35

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

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

cancel g

divide .2

r = 0.75?
 
  • #10
brentwoodbc said:
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:
 
  • #11


tiny-tim said:
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)
 
  • #12


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.
 
  • #13
ah … got it!

that's fine then :smile:
 
  • #14


tiny-tim said:
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:
 
  • #15
brentwoodbc said:
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
brentwoodbc said:
.2gr=-(.122g(.45 -.25)+.5(.25))

that does make it .5 :redface:
 
  • #16


tiny-tim said:
oh, I missed the minus in yourthat does make it .5 :redface:
pfft. I'm so smart lol :cool:

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.
 

What is static equilibrium?

Static equilibrium is a state in which an object is at rest and all forces acting on it are balanced. This means that the net force and net torque on the object are both equal to zero.

What is the mass problem in static equilibrium?

The mass problem in static equilibrium refers to a common physics problem where the mass of an object is unknown and needs to be solved for using principles of static equilibrium.

What are the key equations used in solving static equilibrium mass problems?

The key equations used in solving static equilibrium mass problems are the equations for net force and net torque. These equations state that the sum of all forces and torques acting on the object must equal zero in order for the object to be in static equilibrium.

How do you approach solving a static equilibrium mass problem?

To solve a static equilibrium mass problem, you must first identify all the forces and torques acting on the object and draw a free body diagram. Then, use the equations for net force and net torque to set up a system of equations. Finally, solve for the unknown mass using algebraic techniques.

What are some common mistakes made when solving static equilibrium mass problems?

Some common mistakes made when solving static equilibrium mass problems include forgetting to include all forces and torques acting on the object, using incorrect sign conventions, and not setting up and solving the equations correctly. It is important to double check your work and make sure all forces and torques are accounted for in the problem.

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