R6MechanicsStretch

Friction

Friction opposes relative motion between surfaces in contact. What makes it different from every other force you have met is that it is not a fixed value — it adjusts itself to whatever is needed, up to a maximum, and only reaches that maximum when the object is about to slide.

35 min Forces and Newton’s Laws

The big picture

The governing relation is an $\textbf{inequality}$ , $F \leq \mu R$ , and taking it seriously is what separates a correct solution from a plausible-looking wrong one. A block at rest under a small push has friction exactly matching the push, not $\mu R$ ; only when the push reaches $\mu R$ does the block begin to move. So the equality holds in exactly two situations — when motion is happening, and at the instant it is about to. Recognising which case a question describes is the whole skill, and it is why "on the point of moving" is such a heavily loaded phrase in exam wording.

What you'll be able to do

Use $F \leq \mu R$ and identify when the equality applies
Determine whether a body remains in equilibrium or begins to slide
Solve problems on horizontal surfaces with friction
Account for an angled applied force changing the normal reaction

Friction adjusts itself

Push a heavy box gently and it does not move: friction has matched your push exactly, keeping the resultant force zero. Push harder and friction increases to match, until it reaches its maximum and the box slides.

That maximum is $\mu R$ , where $R$ is the normal reaction and $\mu$ the $\textbf{coefficient of friction}$ — a dimensionless number depending on the two surfaces. So in general $F \leq \mu R$ .

The body is in $\textbf{limiting equilibrium}$ when $F = \mu R$ and it is still at rest: on the point of moving. Once sliding begins, friction stays at $\mu R$ and opposes the direction of motion.

F \leq \mu R

The general relation — an inequality.

F = \mu R

Only when sliding, or on the point of sliding.

Example 1A 10 kg block rests on rough horizontal ground with

\mu = 0.4

. A horizontal force of 20 N is applied. Does it move, and what is the friction? Take

g = 9.8

1Normal reaction:

R = 10 \times 9.8 = 98

2Maximum friction:

\mu R = 0.4 \times 98 = 39.2

3The applied force of 20 N is below that maximum.

AnswerIt does not move. Friction is

20

N — matching the applied force, not

39.2

Tip — If a body is at rest and not on the point of moving, friction equals whatever force it is balancing. Substituting $\mu R$ there is the single most common error in the topic.

Testing whether a body moves

The test has three steps. Find the normal reaction by resolving perpendicular to the surface. Compute the maximum friction $\mu R$ . Compare it with the force tending to cause motion.

If the driving force is less than $\mu R$ , the body stays in equilibrium and friction equals the driving force. If it exceeds $\mu R$ , the body accelerates, and friction takes its maximum value $\mu R$ opposing the motion.

When it does move, the resultant force is the driving force minus $\mu R$ , and $F = ma$ gives the acceleration.

F_{\text{driving}} - \mu R = ma

Once sliding, friction is at its maximum.

Example 1The same 10 kg block with

\mu = 0.4

is now pushed with 50 N horizontally. Find the acceleration.

1Maximum friction is still

39.2

N, and

50 > 39.2

, so it moves.

2Resultant:

50 - 39.2 = 10.8

a = \dfrac{10.8}{10}

Answer

a = 1.08\ \mathrm{m\,s^{-2}}

Why this works

Note that $\mu R$ does not depend on the applied force or on the contact area — only on the surfaces and the normal reaction. Doubling the push does not change the friction once sliding has begun.

When the applied force is at an angle

An angled force changes two things at once. Its horizontal component drives the motion, and its vertical component alters the normal reaction — which changes the maximum friction as well.

For a force $P$ at angle $\theta$ above the horizontal, resolving vertically gives $R = mg - P\sin\theta$ . The upward component partly supports the body, reducing $R$ and therefore reducing $\mu R$ .

That is why pulling at an angle can be more effective than pushing horizontally, even though less of the force acts forwards: the friction opposing you has fallen too. Pushing $\textit{downwards}$ at an angle has the opposite effect, increasing $R$ and making the object harder to move.

R = mg - P\sin\theta

Pulling upwards at an angle reduces

R

R = mg + P\sin\theta

Pushing downwards increases it.

Example 1A 20 kg crate on ground with

\mu = 0.3

is pulled by a force of 80 N at

25^{\circ}

above the horizontal. Determine whether it moves. Take

g = 9.8

1Vertical:

R = 20(9.8) - 80\sin25^{\circ} = 196 - 33.81 = 162.19

2Maximum friction:

0.3 \times 162.19 = 48.66

3Horizontal component:

80\cos25^{\circ} = 72.50

72.50 > 48.66

AnswerIt moves, since the horizontal component exceeds the maximum friction.

Tip — Always resolve vertically to find $R$ before computing $\mu R$ . Using $R = mg$ when a force acts at an angle gives the wrong maximum friction and often the wrong conclusion.

Limiting equilibrium problems

Questions frequently describe a body "on the point of moving" or "in limiting equilibrium". That phrase is the signal that $F = \mu R$ exactly, and it usually supplies the extra equation needed to find an unknown.

Typical unknowns are the coefficient of friction, the minimum force required to move a body, or the maximum force that can be applied without moving it.

Set up the equilibrium equations in both directions, substitute $F = \mu R$ , and solve.

Example 1A 15 kg block is on the point of sliding when a horizontal force of 44.1 N is applied. Find the coefficient of friction. Take

g = 9.8

1On the point of sliding, so friction is at its maximum and equals the applied force:

F = 44.1

R = 15 \times 9.8 = 147

\mu = \dfrac{F}{R} = \dfrac{44.1}{147}

Answer

\mu = 0.3

Why this works

A coefficient of friction is dimensionless — a ratio of two forces — and for most everyday surfaces lies between about 0.1 and 1. A value far outside that range usually indicates an arithmetic slip.

Think like an examiner

Resolve perpendicular to the surface to find $R$ before using $\mu R$ .
Use $F = \mu R$ only when the body is sliding or on the point of sliding — otherwise friction balances the applied force.
State the comparison explicitly when testing for motion: driving force against $\mu R$ .
Watch for an angled force changing $R$ , and hence the maximum friction.

Common misconceptions

Using

F = \mu R

for a body at rest under a small force.

Friction matches the applied force until the maximum is reached. The relation is

F \leq \mu R

Taking

R = mg

when a force acts at an angle.

Resolve vertically: an upward component reduces

R

, a downward component increases it.

Thinking friction depends on the contact area.

In this model it depends only on

\mu

and

R

Applying friction in the direction of motion.

Friction always opposes relative motion or the tendency to move.

Friction

F \leq \mu R

The general relation.

F = \mu R

Sliding or limiting equilibrium.

R = mg \mp P\sin\theta

Angled force changes

R

F_{\text{driving}} - \mu R = ma

Once moving.

Stretch yourself

A 12 kg block rests on rough horizontal ground with $\mu = 0.25$ . A force $P$ is applied at $40^{\circ}$ above the horizontal. Find the value of $P$ that puts the block on the point of moving. Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .

Hint — Both the horizontal and vertical components involve $P$ , so set up both equations and eliminate $R$ .

Reveal solution

SolutionResolving vertically:

R + P\sin40^{\circ} = 12 \times 9.8

, so

R = 117.6 - 0.6428P

. On the point of moving, friction is limiting and balances the horizontal component:

P\cos40^{\circ} = \mu R

, so

0.7660P = 0.25\left(117.6 - 0.6428P\right) = 29.4 - 0.1607P

. Collecting:

0.7660P + 0.1607P = 29.4

, so

0.9267P = 29.4

and

P = 31.7

N (3 s.f.). Note what the angle bought: pulling horizontally would need

P = \mu mg = 0.25 \times 117.6 = 29.4

N, so this angle actually requires slightly

\textit{more}

force. At

40^{\circ}

the loss in horizontal component outweighs the reduction in friction — there is an optimal angle, and it is smaller than this one.

\blacksquare

Questions students ask

When can I use

F = \mu R

Only when the body is sliding, or is on the point of sliding (limiting equilibrium). At rest under a smaller force, friction equals whatever it is balancing.

Does a larger contact area increase friction?

Not in this model. Friction depends only on the coefficient and the normal reaction, which is why a brick slides the same whichever face it rests on.

Why does pulling at an angle sometimes help?

The upward component reduces the normal reaction and therefore the maximum friction. Whether it helps overall depends on whether that reduction outweighs the loss in horizontal component.

Can the coefficient of friction exceed 1?

Yes, though it is unusual. It would mean the maximum friction exceeds the normal reaction, which happens for some rubber and adhesive surfaces.

Key takeaways

Friction adjusts to oppose motion, up to a maximum of $\mu R$ .
The relation is $F \leq \mu R$ ; equality holds only when sliding or on the point of sliding.
Find $R$ by resolving perpendicular to the surface before computing $\mu R$ .
An angled force changes $R$ and therefore the maximum friction.
"On the point of moving" signals limiting equilibrium and supplies an extra equation.