Friction and Newton's second law

The idea

Static equilibrium is the case where the forces balance. When they don’t, the body accelerates, and Newton’s second law tells you how much:

$\boxed{\sum \mathbf{F}=m\mathbf{a}}$

Written in components, it’s the equilibrium equations with $ma$ on the right instead of zero:

$\sum F_x=m{a}_x=m\ddot{x}\sum F_y=m{a}_y=m\ddot{y}$

Friction appears in almost every real problem, because it’s what a robot’s wheels push against, and what a gripper relies on to hold things.

Friction

Friction acts along the contact surface, opposing the sliding (or the tendency to slide).

	Formula	When
Static	$F_s\le {\mu }_{s}N$	Not sliding. It’s only as big as it needs to be, up to the limit ${\mu }_{s}N$
Kinetic	$F_k={\mu }_{k}N$	Sliding. A fixed value

• $N$ is the normal reaction. Find it from the force balance perpendicular to the surface. It’s only equal to $mg$ on flat ground with no other vertical forces. On a slope, $N=mg\cos \theta$.
• Usually ${\mu }_k<{\mu }_s$: it takes more force to start sliding than to keep sliding. That’s one reason a mechanism’s start-up effort is higher than its running effort.
• Will it slip? Work out the friction needed to stay put and compare it with ${\mu }_{s}N$. If it needs more than that, it slides, and you switch to ${\mu }_k$.

Method for dynamics problems

1. Draw a free body diagram with every force, and choose axes along and perpendicular to the motion (along the slope, for example).
2. Draw the kinetic diagram next to it: an arrow $m\mathbf{a}$ in the direction of acceleration. The FBD and the kinetic diagram must be equivalent.
3. Write $\sum F=ma$ in each direction, for each body.
4. Add the constraints: bodies joined by a taut, inextensible rope share the same acceleration size; friction is $\mu N$.
5. Solve, then sense-check the direction and size.

Worked example: a block on a slope

A 12 kg block sits on a 25° slope with ${\mu }_s=0.3$ and ${\mu }_k=0.2$. Does it slide? If so, find its acceleration.

Answer

Perpendicular to the slope (no acceleration that way): $N=mg\cos 25°=12\times 9.81\times 0.906=106.7$ N Does it slip? To stay still, friction would need to equal $mg\sin 25°=49.8$ N. The maximum available is ${\mu }_{s}N=0.3\times 106.7=32.0$ N, which isn’t enough, so it slides. (Shortcut: it slides whenever $\tan \theta >{\mu }_s$; $\tan 25°=0.466>0.3$.) Along the slope (down = +): $mg\sin 25°-{\mu }_{k}N=ma$ $49.8-0.2\times 106.7=12a$ → $49.8-21.3=12a$ → $a=2.37$ m/s² down the slope Notice the mass cancels if you write it symbolically: $a=g(\sin \theta -{\mu }_k\cos \theta )$.

Worked example: connected bodies

A 4 kg block on a table (${\mu }_k=0.25$) is tied by a light rope over a smooth pulley to a 3 kg mass hanging over the edge. Find the acceleration and the rope tension.

Answer

Same rope, so both move with the same $a$. Take the direction of motion as positive for each body. Hanging mass: $3g-T=3a$ Table block: $T-0.25(4g)=4a$ Add the two equations (T cancels): $3g-g=7a$ → $a=2\times 9.81/7=$ 2.80 m/s² $T=3(9.81-2.80)=$ 21.0 N, which is less than the hanging weight (29.4 N), as it must be for the mass to accelerate downwards.

Rotation: ΣM = Iα

For a body turning about a fixed point, the rotational version of $F=ma$ is

$\boxed{\sum M=I\alpha }$

For a single small mass at radius r, the mass moment of inertia is $I=m{r}^2$. Applying it to a pendulum: the tangential force $F_t$ gives a moment $F_{t}r=m{r}^2\alpha$, so $\alpha =F_t/(mr)$.

Worked example: a pendulum at release

A 1.5 kg ball on a 0.8 m string is held at 20° from vertical and released. At the moment of release, find the angular acceleration and the string tension.

Answer

At release it isn’t moving yet ($\omega =0$), so there’s no normal (centripetal) acceleration. Tangential: $F_t=mg\sin 20°=1.5\times 9.81\times 0.342=5.03$ N → $\alpha =F_t/(mr)=5.03/(1.5\times 0.8)=$ 4.19 rad/s² Along the string: $T-mg\cos 20°=m{\omega }^{2}r=0$ → $T=13.8$ N Once it’s swinging, $\omega \ne 0$ and the tension rises by $m{\omega }^{2}r$.

Why it matters for robots

• Wheel traction: a robot can’t accelerate harder than friction allows: $a_{max}={\mu }_{s}g$ on flat ground. Push the motor beyond that and the wheels spin.
• Motor sizing: the force a drive must supply is $ma$ plus friction or rolling resistance plus any slope. The same sum appears in Sizing a drive motor.
• Mechanisms: a slider-crank’s joint forces come from $\sum F=ma$ on the slider and $\sum M=I\alpha$ on the rod.

Common mistakes

• Taking $N=mg$ on a slope, or when another force pushes down or pulls up.
• Using ${\mu }_s$ for something that’s already sliding.
• Friction in the wrong direction. It always opposes the sliding, so check which way the body would move without it.
• One equation for two bodies. Each body gets its own $\sum F=ma$, linked by the rope tension.

Check yourself

1. A 20 kg crate on the floor ( ${\mu }_k=0.3$) is pushed horizontally with 100 N. Find its acceleration.

$N=196.2$ N, friction $=58.9$ N. $a=(100-58.9)/20=$ 2.06 m/s².

2. What's the steepest slope a robot can climb at constant speed if its wheels have ${\mu }_s=0.6$ (ignore rolling resistance)?

Grip runs out when $\tan \theta ={\mu }_s$: $\theta ={\tan }^{-1}(0.6)=$ 31°.

3. Why does a mechanism need more torque to start than to keep running?

Static friction is higher than kinetic friction, the motor has to accelerate all the moving masses from rest ($\sum F=ma$, $\sum M=I\alpha$), and at some positions (a slider-crank near dead centre) the crank has little leverage.

Next: where the weight actually acts, in Centre of mass.