Confirm this problem which involves friction

[danielI]

God day!

The magnitude of force P is slowly increased. Does the homogeneous box of mass m slip or tip first? State the value of P which would cause each occurence. Neglect any effect of the size of the small feet.

http://img75.imageshack.us/img75/629/friction5mb.png

This is my work which I wish anyone could check.

N + P_y - mg = 0\Rightarrow N = mg - P_y

We also know that P_y = P\sin30\Rightarrow N = mg - P\sin30

The resultant force of the forces that will affect the body in the y-axis is
R_y = mg - P\sin30

And in the x-axis it will be the force that is pulling the body minus the friction, i.e.,

R_x = P\cos30 - \frac{mg-P\sin30}{2}

Now is everything correct? Could I have missed some force or misscalculated something?

My strategy is now to check when R_x = 0 and R_y = 0, that is, before the storm breaks loose.

R_y = 0 for P = 2mg
R_x = 0 for P = \frac{4mg}{4\cos30+1}

Since \frac{4mg}{4\cos30+1}\leq 2mg it will be starting to move in x-direction before tilting. And it will do this for P > 2mg

Thank you and have a god day!

 

[durt]

You'll have to consider the torques when it tilts.

 

[danielI]

Hello durt

Okay, since the tilting part was wrong we keep the R_x and remove the R_y.

Let the blue dot be origo. We will have two torques, the one by gravity and the one created by the force P. We start of with the one created by gravity.

M_1 = mgd

This will be counterclockwise. Then we look at the other one

M_2 = P_yz

we know that x = d/\tan30. This gives z = (2d + d/\tan30)\sin30 = (d + d/(2\tan30))
So,
M_2 = P(d + d/(2\tan30))

Which is clockwise. We set M_1=M_2 and get P = \frac{mg}{1+\frac{1}{2tan30}}

Then the body will tilt first if \frac{1}{1+\frac{1}{2tan30}} < \frac{1}{\cos30 + 1/4} which clearly is true (well, if you use the calculator ). And hence, the body will tilt when P > \frac{mg}{1+\frac{1}{2tan30}}