Given:
Diameter, d = 20 cm
Height, h = 40 cm
Mass, m = 10 kg
Given that the surface is slowly inclined to the horizontal, let's find the following:
• (A) Show the forces acting on the block in a labelled sketch when the surface is tilted. Treat the cylinder as a point.
The forces acting on the block are sketched below:
• (B) What is the maximum angle the rough surface can make with the horizontal before the cylinder topples over give answer to nearest degree.
The normal force (N) = 0 at the time the cylinder topples.
Before the the cylinder topples, tnet = 0
Apply the formula:
[tex]mg\sin \theta\ast\frac{h}{2}-mg\cos \theta\ast\frac{d}{2}=0[/tex]
Rewrite the equation for θ:
[tex]\begin{gathered} \sin \theta h=\cos \theta d \\ \\ \frac{\sin\theta}{\cos\theta}=\frac{d}{h} \\ \\ \tan \theta=\frac{d}{h} \\ \\ \theta=\tan ^{-1}(\frac{d}{h}) \end{gathered}[/tex]
Where:
d = 20 cm
h = 40 cm
We have:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{20}{40}) \\ \\ \theta=\tan ^{-1}(0.5) \\ \\ \theta=26.57\approx27^0 \end{gathered}[/tex]
Therefore, the maximum angle the rough surface can make with the horizontal before the cylinder topples over is approximately 27 degrees.
