Torque and Equillibrium problem.

  • Thread starter Atomos
  • Start date
  • #1
159
0
A uniform board is leaning against a wall. The board has an angle theta with respect to tge horizontal. If the coefficient of friction between the ground and the board and the wall and the board is 0.650, what is the minimum value of theta such that the board can rest between the two surfaces?

I produced a diagram, and concluded that the net torque about the bottom of the board is 0. The torque forces I included were the torque from the centre of gravity, the torque from the wall, and the torque induced by friction on the wall. I went through and ended up with a trigonometric equation in two variables, theta and mass, that I could not cancel. I am not even going to bother to show my work because my logic must be wrong.

Any suggestions?
 

Answers and Replies

  • #2
Pyrrhus
Homework Helper
2,178
1
Try a Sum of torques about the point the board meets the wall. :wink: and consider [itex] \sum F_{y} = N_{ground-ladder} - mg = 0 [/itex]. Sorry this is not applicable i misread the problem.
 
Last edited:
  • #3
159
0
ERRRRRRRR
I totally misread the question; there is no friction with the wall. So I can solve it easily.

With regards to a problem where both the wall and the wall of friction:
That summation of vertical forces, it does not include the upwards force of friction (if the ladder is in motion) exterted on the ladder by the wall, or is that completely irrelevent?
 
  • #4
Pyrrhus
Homework Helper
2,178
1
Ok Let's see what you got

[tex] \sum F_{x} = N_{wall} - F_{friction/ground} = 0 [/tex]

[tex] \sum F_{y} = N_{ground} + F_{friction/Wall} - mg = 0 [/tex]

[tex] \sum \tau_{ground} = N_{wall} \sin \theta + F_{friction/Wall} \cos \theta - mg \frac{1}{2} \cos \theta = 0 [/tex]

Right?
 
Last edited:
  • #5
Pyrrhus
Homework Helper
2,178
1
Now consider the following relationship

[tex] \sum \tau_{ground} = N_{wall} \sin \theta + \mu N_{wall} \cos \theta - mg \frac{1}{2} \cos \theta = 0 [/tex]

[tex] N_{wall} = \mu N_{ground} [/tex]

[tex] N_{wall} = \mu (mg - F_{friction/Wall}) [/tex]

[tex] N_{wall} = \mu (mg - \mu N_{wall}) [/tex]

[tex] N_{wall} = \frac{\mu mg}{1 + \mu^2} [/tex]
 
Last edited:
  • #6
159
0
in the second equation, I think normal from ground and friction from wall should be of the same sign, and with the third one, I think normal torque from wall and torque from friction should also be of the same sign.
 
  • #7
Pyrrhus
Homework Helper
2,178
1
Ok i think is all fixed now. Any more quibbles? :approve:
 

Related Threads on Torque and Equillibrium problem.

  • Last Post
Replies
2
Views
479
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
897
Replies
8
Views
6K
  • Last Post
Replies
8
Views
1K
Top