Homework Help: Force Equilibrium Problem

1. Dec 29, 2012

has1993

1. The problem statement, all variables and given/known data
A pole is suspended horizontally between a wall and an attached string.(See the attachment). An object(W) is hanging from the pole at a distance of x from the wall. Find the value of x that preserves the equilibrium.

1) the angle between the pole and the string(q) = 10°
2) the mass of the object(W) = mass of the pole
3) static friction coefficient between the wall and the pole = 0.3
4) the maximum angle at which the pole could be kept in horizontal (without the object) = 16.7°

2. Relevant equations
torque = distance × force

3. The attempt at a solution

I've tried to build a linear system of 3 equations for the pole without the object by taking the torques at the wall-end, string-end and half-way to the center of gravity of the pole (at l/4) so:

W*l/2 = T * l * Sin(16.7) -------> 1
W*l/2 = l * T*0.3*Cos(16.7) --------->2
l*T*0.3*Cos(16.7)/4 + W*l/4 = 3*l*T*Sin(16.7)/4 --------->3

l = length of the pole

By solving the system i tried to find the values of l, and W so i can use them to solve the one with the object. I don't know even if this is correct but i can't find a way to solve it if it's correct. So any help would be really appreciated. (And sorry for my coarse English. This is actually the first time i've posted in a thread)

2. Dec 29, 2012

SteamKing

Staff Emeritus
No attachment

3. Dec 29, 2012

SteamKing

Staff Emeritus
You should write the standard equations of equilibrium, including the reactions which the pole imposes on the wall and the friction force generated. The 3 equations you have developed so far are just a portion of what is required to determine equilibrium.

4. Dec 29, 2012

Simon Bridge

No attachment ... something like this:

... only your object is hanging from a distance x from the wall, not the end?
if the length of the pole is $L$, $0<x<L$.

1) the angle between the pole and the string(q) = 10°
The string angle is $\theta$ - the tension $T$ in the string will produce components along the pole and perpendicular to it.

2) the mass of the object (W) = mass of the pole
... helpful to use $W$ for the weight of the object - let it's mass be $M$ so $W=Mg$.

3) static friction coefficient between the wall and the pole = 0.3
... friction force: $f=\mu T\cos(\theta)$

4) the maximum angle at which the pole could be kept in horizontal (without the object) = 16.7°
... this tells you that $f+T\sin(16.7^\circ)=W$ and $f=T\sin(16.7^\circ)$.

Notice that the end by the wall could slip so I wouldn't treat it as a fixed pivot... so you should draw your force diagram as a free rod.

5. Dec 30, 2012

has1993

oops.. my bad here's the attachment

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6. Jan 1, 2013

Simon Bridge

https://www.physicsforums.com/attachment.php?attachmentid=54365&d=1356930096

Let the length of the pole be L so it's com is L/2
I believe my comments still stand.

7. Jan 3, 2013

has1993

Sure it still stands. So can you please clarify it a little bit more on finding the value of 'x'. (It would immensely help if you can show me how to build the equations necessary). I'm hoping that I'm not overly bugging you.

8. Jan 3, 2013

haruspex

There's no point in the developing the third torque equation. It must be deducible from the first two. And all you can deduce from the first two equations is that tan(16.7 degrees) = 0.3. In short, fact (4) of the OP isn't telling you anything extra. There's no benefit in analysing the case without the load.
Please post your equations for the case with the load, and however far you get with solving them.

9. Jan 3, 2013

Simon Bridge

I'm sure it would - that would amount to "doing the problem for you". I know it's a pain, but you learn more if you do it yourself.

The effect of weight W is to apply an extra perpendicular force to the rod at position (L/2)-x from it's center of mass ;)

I'd start by drawing a diagram with just the rod and all the forces on it ... linear forces must all cancel (since the rod does not go anywhere) and the torques must all cancel (because it is not turning) and they are all related to each other.

10. Jan 4, 2013

has1993

Okay I think I've made some progress

Writing for the linear forces I got this:

2W = mu * Tcos(10) + Tsin(10) ------> 1

So

W/T = -0.4

Then I took torques at the wall-end :

WL/2 + Wx = LTSin(10) -------> 2

And substituted W = -0.4T in the second equation and thus

L/x = 1.18

But I have no clue as to what I should do next.

11. Jan 4, 2013

haruspex

I get W/T = 0.2345. It's 10 degrees, not 10 radians!
With the corrected W/T and taking sin(10*pi/180 radians) you should get x/L = 0.24.
Now, the OP says "Find the value of x that preserves the equilibrium". Ther calculations assumed the frictional force was the maximum possible, so the x/L computed is the maximum. I.e. any value of x less than this will still be in equilibrium.

12. Jan 4, 2013

Simon Bridge

I though their point (4) was telling us the minimum friction that would stop the wall-end from sliding - with no added load?

Wouldn't we expect the smaller angle to indicate a larger friction though possibly a smaller perpendicular force?

x is measured from the wall-end.
doesn't the load on the wall end increase as x decreases?
doesn't that mean it is possible that the static friction could be overcome by the load for small x?

13. Jan 4, 2013

haruspex

As I posted previously, item (4) adds nothing. All it provides is the tangent of 10 degrees.
But, yes, I did get it backwards. I should have said that the value of x arrived at is the minimum value, not the maximum. Thanks.

14. Jan 23, 2013

has1993

Finally I got it figured. I had overlooked the original problem statement that it was a meter ruler hanging from the string. So l = 1m. And I felt kinda stupid when i found that. But huge thanks goes to everyone who had to waste their valueble time for my stupidity. :D

15. Jan 23, 2013

Simon Bridge

Well done.
You should not be scared of looking stupid - we've all been there :)