Friction and applications of Newton's laws

Last week we built Newton’s laws in a frictionless world. In real questions there is almost always friction, and the questions on the midterm that need the most calculation come from this week: inclined plane, angled pull, table-pulley, Atwood. The solution method is the same for all of them. Free-body diagram, choice of axes, $\sum F = ma$ on each axis. The only new piece on this page is getting the magnitude and direction of friction right.

Golden rules:

1. Friction is $\mu N$, not $\mu mg$. Find $N$ first.
2. First decide static or kinetic. Is the object sliding or not?
3. Static friction is as much as needed. $\mu_s N$ is only an upper limit.
4. Friction opposes sliding. Not the direction of motion.
5. For connected objects, a separate diagram for each object. The same rope, the same $T$, the same $|a|$.

Static and kinetic friction

If two surfaces are not sliding relative to each other, the force between them is static friction; if they are sliding, it is kinetic friction.

$f_s \le \mu_s N \qquad\qquad f_k = \mu_k N$

$\mu_s$ and $\mu_k$ are the dimensionless coefficients of friction. Usually $\mu_k < \mu_s$: it is harder to start a crate moving than to keep it moving.

Static friction is an inequality. If you push lightly on a book on a table, friction is exactly equal to the push and the book does not move. As you increase the push, friction increases too. $f_s$ can be at most $\mu_s N$; the moment this value is exceeded, the object starts to slide and friction drops to $\mu_k N$.

Example 1. A 30 kg crate, $\mu_s = 0.40$, $\mu_k = 0.30$. (a) 100 N and (b) 150 N horizontal push.

N = mg = 30 · 9.80 = 294 N
f_s,max = 0.40 · 294 = 117.6 N
(a) 100 N < 117.6 N → does not move, f_s = 100 N, a = 0
(b) 150 N > 117.6 N → slides, f_k = 0.30 · 294 = 88.2 N
    a = (150 - 88.2) / 30 = 2.06 m/s²

PITFALL · writing μs N for an object at rest. In (a) the friction is 100 N, not 117.6 N. If friction were greater than the push, the crate would accelerate toward you. Static friction equals $\mu_s N$ only at the threshold of sliding.

Direction of friction

Friction opposes the sliding between the two surfaces, or the tendency to slide. When you walk, your foot pushes the floor backward; the floor applies static friction to the foot in the forward direction. The same for a car: the wheel pushes the road backward, the road pushes the car forward. That is why the sentence “friction is always opposite to the motion” is a trap on the exam.

The normal force is not always mg

Since friction depends on $N$, if you get $N$ wrong the whole question goes off. If there is no acceleration on the perpendicular axis, write $\sum F_y = 0$ and get $N$ from there:

Example 2: Angled pull. A 15 kg sled is pulled with an 80 N force 25° above the horizontal, $\mu_k = 0.35$.

Vertical: N + 80 sin25° - 15 · 9.80 = 0
          N = 147 - 33.8 = 113.2 N
f_k = 0.35 · 113.2 = 39.6 N
Horizontal: 80 cos25° - 39.6 = 15 a
            72.5 - 39.6 = 15 a  →  a = 2.19 m/s²

Pulling upward at an angle lifts the object slightly, reducing the normal force and the friction. That is why it is easier to pull a snow sled with the rope angled upward.

Friction on an incline

Axes: $x$ along the incline, $y$ perpendicular to the surface. $N = mg\cos\theta$, friction $\mu_k mg\cos\theta$.

Example 3. On a 30° incline with $\mu_k = 0.20$, the block slides down.

a = 9.80 (sin30° - 0.20 cos30°)
  = 9.80 (0.500 - 0.173) = 3.20 m/s²

For an object thrown up, both the component of the weight and friction are directed downward, and the slowing is greater. After it stops at the top, if $\mu_s < \tan\theta$ it starts to slide back, and this time friction turns upward.

Example 4: Does it slide or not? A 5.0 kg block is placed on a ramp with $\mu_s = 0.50$ and $\mu_k = 0.40$. What happens (a) when the ramp is at 25°, (b) at 30°?

Condition for sliding: mg sinθ > μs mg cosθ  ⇔  tanθ > μs
(a) tan25° = 0.466 < 0.50 → the block STAYS
    Static friction as much as needed: f_s = mg sin25° = 5.0 · 9.80 · 0.423 = 20.7 N
    (The upper limit μs mg cos25° = 22.2 N is not exceeded.)
(b) tan30° = 0.577 > 0.50 → the block SLIDES, now μk applies
    a = g (sin30° - 0.40 cos30°) = 9.80 (0.500 - 0.346) = 1.50 m/s²

Writing 22.2 N as the friction force in (a) is the incline version of the trap in Example 1. For a block at rest, friction is exactly as much as balances the component of the weight along the incline.

TIP · limit test. Put $\mu_k = 0$: you should get $g\sin\theta$. For sliding down with friction the answer should be smaller than $g\sin\theta$, for sliding up larger. Half the options are eliminated with this check.

Connected objects

Table + hanging mass

$m_1$ is on the table (friction $\mu_k$), $m_2$ hangs over the pulley. A separate equation for each object, taking the direction of acceleration as +:

$m_2:\ m_2 g - T = m_2 a \qquad m_1:\ T - \mu_k m_1 g = m_1 a$

Add them, and $T$ cancels:

$a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2}$

Example 5. $m_1 = 4.0$ kg, $\mu_k = 0.25$, $m_2 = 2.0$ kg.

f_k = 0.25 · 4.0 · 9.80 = 9.80 N
a = (19.6 - 9.80) / 6.0 = 1.63 m/s²
T = m2 (g - a) = 2.0 (9.80 - 1.63) = 16.3 N
Check: T = m1 a + f_k = 6.53 + 9.80 = 16.3 N

If $m_2 g < \mu_s m_1 g$, the system never starts moving; check this first.

Atwood machine

Two masses hang from a frictionless pulley ($m_2 > m_1$):

$a = \frac{(m_2 - m_1)g}{m_1+m_2},\qquad T = \frac{2m_1m_2 g}{m_1+m_2}$

The tension lies between the two weights: $m_1 g < T < m_2 g$. If the rope fully supported the heavy object, the system would not accelerate.

PITFALL · tension = hanging weight. In an accelerating system $T \ne m_2 g$. If the hanging object accelerates downward, the rope holds it with less force than its weight: $T = m_2(g - a)$.

Roadmap for connected-object questions

1. Draw a separate free-body diagram for each object. Take the pulley as light and frictionless: the tension is the same at both ends of the rope.
2. For each object choose “direction of motion is +”. For example, right is + for the block on the table and down is + for the hanging block. That way the accelerations of the two objects are the same signed $a$.
3. Add the equations. $T$, an internal force, cancels and $a$ is left alone. The denominator has the total mass, and the numerator has the net external force that moves the system.
4. Get $T$ from the simplest equation, and check it with the other equation.

This pattern fits every setup: for a block on an incline + hanging mass, the numerator is $m_2 g - m_1 g\sin\theta - \mu_k m_1 g\cos\theta$ and the denominator is again $m_1 + m_2$. If the result comes out negative, the direction you assumed is wrong; the system either goes the other way or does not move at all because of static friction.

Common mistakes