Need help formulating moments about a point.

[rock.freak667]

Homework Statement

http://img376.imageshack.us/img376/1033/questionii7.jpg

The diagram shows a cross-section ABCD of a uniform rectangular block of weight W. The lengths of AB and BC are 2a and a respectively. The edge through A rests against a smooth vertical wall and the edge through B rests on a rough horizontal floor. The coefficient of friction between the block and the floor is $\mu$. The block is in equilibrium with AB inclined at an angle $\alpha$ to the vertical. Show that the wall exerts a force of magnitude $\frac{1}{4}(2tan\alpha -1)W$ on the block.

Show also that $tan^{-1}(\frac{1}{2}) \leq \alpha \leq tan^{-1}(\frac{1}{2}+2\mu)$

Homework Equations

$$\tau = \vec{F}\times \vec{r}=Frsin\theta$$

The Attempt at a Solution

In order to not have friction included, I decided to take moments about B.

The distance of W from B is $\frac{\sqrt{5}a}{2}$ (I think I did that correctly).
The normal reaction at A, R, acts parallel to surface on which B lies.

So the clockwise moment is R*2a and the anti-clockwise moment is $\frac{\sqrt{5}a}{2}W$

is this correct so far, because I don't think I formulated the distances correctly.

 

[sirzerp]

Well if the wall is smooth then all the weight presssing against it is in affect pressing hortizonal. In this case the weight is either pressing down on point B or action-reaction against point A.

Prorating the weight shared by points B and A, at what angle does the force of A overcome the mu of B?

 

[rock.freak667]

Prorating the weight shared by points B and A, at what angle does the force of A overcome the mu of B?

Not too sure on what you mean here. The normal reaction at is horizontal, that I get. The frictional force,at B acts in the opposite direction of the normal reaction at A. If the body is in equilibrium, doesn't that mean that the normal reaction at A=Frictional force?

 

[sirzerp]

Not too sure on what you mean here. The normal reaction at is horizontal, that I get. The frictional force,at B acts in the opposite direction of the normal reaction at A. If the body is in equilibrium, doesn't that mean that the normal reaction at A=Frictional force?

But the Frictional force depends on the normal force which depends on the amount of weight which depends on the angle.

Simple. When you decrease the normal force you are increasing the horizontal force. At some point for some mu, the horizontal force will win.

Finding the math for all of this, might be a bit harder.

 

[rock.freak667]

Well usually the normal reaction is mgcos$\alpha$. I can split W into sine and cosine. But that doesn't seem to help.

 

[alphysicist]

Hi rock.freak667,

The Attempt at a Solution

In order to not have friction included, I decided to take moments about B.

The distance of W from B is $\frac{\sqrt{5}a}{2}$ (I think I did that correctly).
The normal reaction at A, R, acts parallel to surface on which B lies.

So the clockwise moment is R*2a and the anti-clockwise moment is $\frac{\sqrt{5}a}{2}W$

is this correct so far, because I don't think I formulated the distances correctly.

Your distances are okay, but the moments are not correct. The torque is not F times r; you have to account for the fact that the torque only takes into account the perpendicular parts of F and r. So to use the distances you found you would need to find the angle between W and the diagonal and use a trig function for the weight torque (and a similar thing for the wall's normal force torque).

However, there is an easier approach. Since W is a vertical force, you just need to find the horizontal distance from the pivot to the center of gravity (the horizontal component of the displacement from pivot to center). Then W times this horizontal distance is the torque magnitude.

Similarly, since the normal force at A is horizontal, you just need to find the vertical distance from the pivot to the point A (which is the same as the distance from the origin to point A); multiplying these together gives the torque magnitude from the wall's normal force.

Once you find these distances and therefore the torques, setting these magnitudes equal to each other (like you mentioned in your first post) should give the desired answer for the first part. What do you get?