Question

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?

Answer 1

15.38°

Explanation:

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:

`\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 \\ (\sin \theta)/(\cos \theta)=(d_1)/(d_2) \\ \tan \theta=(d_1)/(d_2) \\ d_1=(1.1)/(2)=0.55\text{ m} \\ d_2=2\text{ m} \\ \text{ replacing} \\ \tan \theta=(0.55)/(2) \\ \theta=\tan ^(-1)((0.55)/(2)) \\ \theta=15.38\text{\degree} \end{gathered}`

The angle is 15.38°