# Mechanics problem

• Oct 30th 2006, 12:25 PM
aphw
Mechanics problem
I've been working on this problem but I can't finish it off.

Here's what I've done so far. As you can see, whilst trying to work out (a) I've accidentally worked out (b). But I can't answer (a)! There just doesn't seem to be enough information and I end up going round in circles trying to find R (on my diagram).
:confused:

• Oct 30th 2006, 01:10 PM
topsquark
Quote:

Originally Posted by aphw
I've been working on this problem but I can't finish it off.

Here's what I've done so far. As you can see, whilst trying to work out (a) I've accidentally worked out (b). But I can't answer (a)! There just doesn't seem to be enough information and I end up going round in circles trying to find R (on my diagram).
:confused:

Edit: Whoops! My post for part a) got lost. I'll post that in a minute.

b) Again using your notation. I'm choosing a +x direction in the direction of the vector X and a +y direction in the direction of +y.

As you correctly noted:
$\displaystyle \sum F_x = X - T = 0$
$\displaystyle \sum F_y = Y - W + R = 0$

So X = T = (1/4)W and Y = W - R = W - (7/8)W = (1/8)W using the results of part a.

Now, Y is a static friction force which is only as large as it needs to be to hold the ladder up. If I assume that static friction is at its maximum, then the value of $\displaystyle \mu$ that I calculate would be the minimum value we would require to hold the ladder up. (Note: X is the normal force at this point.)

So:
$\displaystyle Y = \mu X = \mu (1/4)W$

Thus:
$\displaystyle Y = \mu (1/4)W = (1/8)W$ from above.

Solving this for $\displaystyle \mu$ gives a value of 1/2. Thus
$\displaystyle \mu \geq 1/2$

-Dan
• Oct 30th 2006, 01:19 PM
topsquark
Quote:

Originally Posted by aphw
I've been working on this problem but I can't finish it off.

Here's what I've done so far. As you can see, whilst trying to work out (a) I've accidentally worked out (b). But I can't answer (a)! There just doesn't seem to be enough information and I end up going round in circles trying to find R (on my diagram).
:confused:

a) Using your notation we are looking for R.

The nice thing about static equilibrium problems is that we can choose any darned point we like for the axis of rotation. Here's a hint: If you have one or more unknown forces acting on a point, this point is typically a good candidate for an axis of rotation.

In this case note that we have 3 unknown forces: X, Y, and R. If we choose the axis of rotation to be at the point where X and Y are both acting, then the resulting torque equation will be only in terms of R. So my axis of rotation here is going to be the point where the ladder meets the wall. I am going to choose a CCW torque to be in the positive "direction."

$\displaystyle \sum \tau = aWsin(90 - \theta) + (3/2)aTsin(180 - \theta) - 2aRsin(180 - (90 - \theta)) = 0$

Simplifying the trig expressions and noting that T = (1/4)W:
$\displaystyle 0 = aWcos \theta + (3/2)a \cdot (1/4)Wsin \theta - 2aRcos \theta$

Solving this for R gives:
$\displaystyle R = (1/2)W + (3/16)Wtan \theta$

We know that $\displaystyle tan \theta = 2$, so

$\displaystyle R = (7/8)W$

-Dan
• Oct 30th 2006, 01:22 PM
topsquark
Quote:

Originally Posted by aphw
I've been working on this problem but I can't finish it off.

Here's what I've done so far. As you can see, whilst trying to work out (a) I've accidentally worked out (b). But I can't answer (a)! There just doesn't seem to be enough information and I end up going round in circles trying to find R (on my diagram).
:confused:

c) If you look carefully the only part of the solution we used the torque equation was in part a). Since $\displaystyle \sum \tau = I \alpha$ we technically need to know what the object is in order to supply the correct expression for the moment of inertia, I. However in the torque equation we used, the angular acceleration is 0 rad/s^2, so the value of I was immaterial.

Conclusion: I have no idea why we would need to know we are modelling the ladder as a rod. This fact was never used to do these calculations.

If your professor disagrees with me, I would be interested in knowing his/her logic.

-Dan
• Oct 30th 2006, 01:45 PM
aphw
Thanks very much.

So it's not possible to get a physical value for R? Can it only be found in terms of W?
• Oct 30th 2006, 01:49 PM
topsquark
Quote:

Originally Posted by aphw
Thanks very much.

So it's not possible to get a physical value for R? Can it only be found in terms of W?

The answer is that it can only be found in terms of W. Think of it this way, if the ladder were made of styrofoam R would be much less than if it were made of lead. (Whereas Y would change some, but would max out at some point, so it wouldn't be able to support much of the weight.) So this is a sensible result.

-Dan