Lesson 3 · 40 min
Rectangular Coordinates: Friction and Inclines
When a particle moves along a straight line, or its path is easy to describe in \(x\) and \(y\), the equation of motion splits into rectangular components. Most of the work is then in two places: choosing axes that make the acceleration simple, and getting the friction right. Friction is the force that most often decides whether something moves at all.
Learning objectives
- Write the equations of motion in rectangular components, with axes chosen along and across the motion.
- Apply the laws of dry friction: \(F_f \le \mu_s N\) without slipping, \(F_f = \mu_k N\) while slipping.
- Decide whether a body slips by finding the friction it needs and comparing it with the friction available.
- Analyze a block on a rough incline, and a body pushed or pulled at an angle, where \(N \ne mg\).
The equations of motion in components
Resolve each force, and the acceleration, into components along fixed \(x\), \(y\) and \(z\) axes. The vector equation \(\sum\Fvec = m\avec\) becomes three scalar ones:
Rectangular components
\[ \textstyle\sum F_x = m\colMA{a_x} \qquad \sum F_y = m\colMA{a_y} \qquad \sum F_z = m\colMA{a_z} \]- For motion in a plane, two equations are enough.
- On a straight path, put one axis along the motion and the other across it. Across the path the acceleration is zero, so that equation is an equilibrium equation, and it usually gives the normal force.
- The axes do not have to be horizontal and vertical. On an incline, axes along and normal to the slope are almost always best: the acceleration then has only one component.
Dry friction
Where two dry surfaces touch, the contact force has two parts: the normal force \(N\), perpendicular to the surfaces, and the friction force \(\colF{F_f}\), along them. Friction opposes slipping, and how large it is depends on whether the surfaces slip.
Coulomb friction
\[ \text{No slipping: } \colF{F_f} \le \mu_s N \qquad\qquad \text{Slipping: } \colF{F_f} = \mu_k N \]Without slipping, \(F_f\) is whatever the equations of motion need, up to the limit \(\mu_s N\), in whichever direction they need it. While the surfaces slip, \(F_f = \mu_k N\) and points against the motion of one surface relative to the other. Usually \(\mu_k\) is a little smaller than \(\mu_s\).
So there are two cases, and the first step is always to decide which one you are in.
Example 3.1 — A crate on a truck bed
An \(80\ \text{kg}\) crate sits on the flat bed of a truck. Between them, \(\mu_s = 0.4\) and \(\mu_k = 0.3\). The truck accelerates forward at (a) \(3\ \text{m/s}^2\), (b) \(5\ \text{m/s}^2\). Does the crate slip, and what is its acceleration?
Show solution
The only horizontal force on the crate is friction from the bed; it is what drives the crate forward. Vertically, \(N = mg = 784.8\ \text{N}\), so the friction available is \(\mu_s N = 313.9\ \text{N}\).
(a) Assume no slip, so the crate moves with the truck, \(a = 3\ \text{m/s}^2\):
\[ \textstyle\sum F_x = ma: \quad F_f = (80)(3) = 240\ \text{N} \;\le\; 313.9\ \text{N} \]The assumption holds: the crate rides with the truck, at \(3\ \text{m/s}^2\).
(b) Without slip it would need \(F_f = 80(5) = 400\ \text{N} > 313.9\ \text{N}\), so it slips. Then \(F_f = \mu_k N = 235.4\ \text{N}\), still forward, because the bed slides forward under the crate:
\[ a = \frac{235.4}{80} = 2.94\ \text{m/s}^2 \]The crate still speeds up, but more slowly than the truck: relative to the bed it slides back at \(5 - 2.94 = 2.06\ \text{m/s}^2\). The largest acceleration the truck can have without the crate slipping is \(\mu_s g = 3.92\ \text{m/s}^2\), whatever the crate's mass.
A block on a rough incline
A block of mass \(m\) rests on a slope at angle \(\theta\). Before you open Figure 3.1, make a prediction.
Take \(x\) down the slope and \(y\) normal to it. The weight has components \(mg\sin\theta\) down the slope and \(mg\cos\theta\) into it. Across the slope there is no acceleration:
\[ \textstyle\sum F_y = 0: \quad N - mg\cos\theta = 0 \quad\Rightarrow\quad N = mg\cos\theta \]Along the slope, while the block slides down, friction \(\mu_k N\) points up the slope:
Sliding down a rough incline
\[ \textstyle\sum F_x = m\colMA{a}: \quad \colW{mg\sin\theta} - \colF{\mu_k\, mg\cos\theta} = m\colMA{a} \quad\Rightarrow\quad \colMA{a} = g(\sin\theta - \mu_k\cos\theta) \]The block starts to slide only if the weight's component along the slope beats the largest static friction: \(mg\sin\theta > \mu_s\, mg\cos\theta\), or \(\tan\theta > \mu_s\). The mass cancels in both results.
The angle \(\theta = \tan^{-1}\mu_s\) is the angle of repose. Gravel, sand and grain piled any steeper slide until the slope comes back to it, and tilting a board until a block just slips is a simple way to measure \(\mu_s\).
Example 3.2 — Released on a ramp
A \(12\ \text{kg}\) crate is released from rest on a \(25^\circ\) ramp. Between them, \(\mu_s = 0.4\) and \(\mu_k = 0.3\). Find its acceleration and its speed after it has slid \(3\ \text{m}\).
Show solution
Does it slip? \(\tan 25^\circ = 0.466 > \mu_s = 0.4\), so static friction cannot hold it. With \(x\) down the slope:
\[ \textstyle\sum F_y = 0: \quad N = 12(9.81)\cos 25^\circ = 106.7\ \text{N} \] \[ \textstyle\sum F_x = ma: \quad 12(9.81)\sin 25^\circ - 0.3(106.7) = 12a \quad\Rightarrow\quad 49.75 - 32.01 = 12a, \quad a = 1.48\ \text{m/s}^2 \]The acceleration is constant, so \(v^2 = 2as\):
\[ v = \sqrt{2(1.479)(3)} = 2.98\ \text{m/s} \]Forces at an angle: \(N\) is not \(mg\)
The normal force is whatever the equation across the motion makes it, and only on a level floor with nothing else pushing up or down is that \(mg\). Pull a sled with a rope angled upward and the rope carries part of the weight: \(N\) drops, and so does the friction. Push a lawn mower down along its handle and \(N\) grows. Always find \(N\) from its own equation before using \(\mu N\).
Example 3.3 — Pushed up a ramp
A worker pushes a \(50\ \text{kg}\) crate up a \(20^\circ\) ramp with a horizontal force of \(400\ \text{N}\). The crate is moving up the ramp, and \(\mu_k = 0.2\). Find its acceleration.
Show solution
Take \(x\) up the slope and \(y\) normal to it. The horizontal push has components \(400\cos 20^\circ = 375.9\ \text{N}\) up the slope and \(400\sin 20^\circ = 136.8\ \text{N}\) into it. The weight, \(490.5\ \text{N}\), has components \(167.8\ \text{N}\) down the slope and \(460.9\ \text{N}\) into it.
\[ \textstyle\sum F_y = 0: \quad N - 460.9 - 136.8 = 0 \quad\Rightarrow\quad N = 597.7\ \text{N} \]The crate moves up the slope, so friction, \(0.2(597.7) = 119.5\ \text{N}\), points down it:
\[ \textstyle\sum F_x = ma: \quad 375.9 - 167.8 - 119.5 = 50a \quad\Rightarrow\quad a = 1.77\ \text{m/s}^2 \]Using \(N = mg\cos\theta = 460.9\ \text{N}\) here would underestimate the friction by more than \(27\ \text{N}\).
Check your understanding
Key takeaways
- \(\sum F_x = ma_x\), \(\sum F_y = ma_y\): pick one axis along the motion, so the equation across it gives \(N\).
- Without slipping, friction is whatever is needed, up to \(\mu_s N\). While slipping, it is \(\mu_k N\), against the relative motion.
- To decide, assume no slip, find the friction needed, and compare it with \(\mu_s N\).
- On an incline, \(a = g(\sin\theta - \mu_k\cos\theta)\) sliding down; a block stays put while \(\tan\theta \le \mu_s\).
- \(N\) comes from its own equation: an angled pull lowers it, an angled push raises it.
- Next, Lesson 4 joins bodies with cords and pulleys.