A sculpture is 4.00 m tall and has its center of gravity located 2.00 m above the center of its base. The base is a square with a side of 1.10 m. To what angle θ can the sculpture be tipped before it falls over?

Just before the sculpture begins to fall, the sculpture's net torque must be zero since the sculpture is in static equilibrium at that instant. Take the torque about point D (at the point where one side of the square base is attached to the surface) and solve:

[tex]\begin{gathered} (m\cdot g\cdot\cos \theta)\cdot d_1=(m\cdot g\cdot\sin \theta)\cdot d_2 \\ \cos \theta\cdot d_1=\sin \theta\cdot d_2 \\ \frac{\sin \theta}{\cos \theta}=\frac{d_1}{d_2} \\ \tan \theta=\frac{d_1}{d_2} \\ d_1=\frac{1.1}{2}=0.55\text{ m} \\ d_2=2\text{ m} \\ \text{ replacing} \\ \tan \theta=\frac{0.55}{2} \\ \theta=\tan ^{-1}(\frac{0.55}{2}) \\ \theta=15.38\text{\degree} \end{gathered}[/tex]

The angle is 15.38°