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 occurrence. 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 $$P > \frac{mg}{1+\frac{1}{2tan30}}$$