Precariously Perched Laptop

Another way to think about it is if the laptop is positioned so that there is no net torque, then the center of mass is located right above the pivot point (gravity, the only force in this situation, would apply no torque since distance is zero). If the pivot is the origin and the table represents the x-axis, then the center of mass of the base is at 0 and the center of mass of the screen is at $\frac{L}{2}+\frac{L}{2}cos60$ or $\frac{3L}{4}$ (we only care about the x-coordinates). Finally, since the base is twice as massive as the screen, the laptop's overall CoM is two times closer to the base's CoM as to the screen's CoM, so it would be a third of the way between the two at $\frac{L}{4}$ . Since $\frac{L}{4}$ from the base's CoM is also $\frac{L}{4}$ from the hinge of the laptop, the answer is $\frac{1}{4}$ or 0.25.

If the laptop is on the verge of tipping, the entire normal reaction force will be at the pivot point. Call the length of the laptop hanging over the edge $\alpha$ . Call the mass of the base $M$ . With 1 torque on the left side and 2 torques on the right side, the torque balance equation about the tipping point is:

$\frac{L-\alpha}{L} Mg \frac{L-\alpha}{2} = \frac{\alpha}{L} Mg \frac{\alpha}{2} + \frac{M}{2} g (\alpha + \frac{L}{2} cos(60^{\circ})) \\ \frac{(L-\alpha)^{2}}{2L} = \frac{\alpha^{2}}{2L} + \frac{1}{2} (\alpha + \frac{L}{4}) \\ (L-\alpha)^{2} = \alpha^{2} + \alpha L + \frac{L^{2}}{4} \\ L^{2} - 2L\alpha + \alpha^{2} = \alpha^{2} + \alpha L + \frac{L^{2}}{4} \\ 3L\alpha = \frac{3L^{2}}{4} \\ \alpha = \frac{L}{4}$

i got it wrong 2 times because of using $\cos 60$ = $\cfrac {\sqrt3}{2}$

good thing abt rigid bodies is that you can split it up if you want to get 2 separate rigid bodies and individually calculate the torque about critical points and get the same answer if u considered only one center of mass

The lid of the laptop extends outwards from the base by a distance of $Lcos(60) = \frac{L}{2}$ . For the purposes of this problem that means it can be considered as being equivalent to being a flat board of length $\frac{L}{2}$ and twice its actual density, which is the same as the density of the base. Which means we can consider the whole laptop to be a flat board of length $\frac{3L}{2}$ and uniform density. The pivot point will be in the middle of this flat board.