Forces on the plane as the angle varies

Problem

Plot $F_\parallel(\theta)$ (component along the plane) and $F_\perp(\theta)$ (normal component) of the weight, for $\theta \in [0^\circ, 90^\circ]$. Identify the intersection point and comment on its physical meaning.

Solution

Decomposing the weight $mg$ with respect to an inclined plane of angle $\theta$, the two components are: $F_\parallel(\theta) = mg\sin\theta, \qquad F_\perp(\theta) = mg\cos\theta$

Behaviour of $F_\parallel = mg\sin\theta$. It starts at $0$ for $\theta = 0^\circ$ (horizontal plane: no driving component) and increases monotonically up to $mg$ for $\theta = 90^\circ$ (vertical plane: the whole weight pushes along the plane). It is an increasing sinusoidal curve.

Behaviour of $F_\perp = mg\cos\theta$. It starts at $mg$ for $\theta = 0^\circ$ (horizontal plane: the whole weight presses on the plane) and decreases to $0$ for $\theta = 90^\circ$. It is a decreasing cosinusoidal curve.

Intersection point. The two curves meet where $\sin\theta = \cos\theta$, that is: $\tan\theta = 1 \quad\Longrightarrow\quad \theta = 45^\circ$ At that point both hold: $F_\parallel = F_\perp = mg\sin 45^\circ = \frac{mg}{\sqrt 2} \approx 0{,}707\, mg$

Physical meaning. At $45^\circ$ the push along the slope and the normal pressure on the plane are equal: half the “geometric energy” of the weight goes in each direction. For $\theta < 45^\circ$ the normal component dominates (the plane is only slightly inclined, it holds the body well); for $\theta > 45^\circ$ the parallel component dominates (the body strongly tends to slide).

$\ev{F_\parallel = mg\sin\theta \nearrow,\; F_\perp = mg\cos\theta \searrow; \; \text{crossing at } \theta = 45^\circ,\; F_\parallel = F_\perp = \tfrac{mg}{\sqrt 2}}$

Linked items

Topics: Dinamica Concepts: Forza peso Skills: Scomposizione sul piano inclinato · Uso della trigonometria Objects: Piano inclinato