Md1–Md5 - Modelling friction

Md1–Md5 - Modelling friction

Md1–Md3 describe normal contact, friction and their resultant, with a model that distinguishes resting from sliding. Md4 derives the condition for impending slipping on a slope, and Md5 applies Newton's laws with friction in the correct direction.

Contact has normal and tangential components

Would doubling a block's mass make it start slipping down the same rough slope at a smaller angle? More weight pulls it downhill, but more weight also presses it against the surface. We will find which effect wins in this friction model.

You already know

A body's weight is mgmg, vertically downwards. Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}} unless told otherwise. Newton's second law relates the resultant force on one body to its acceleration: ∑F=ma\sum F=ma along a chosen direction.

Contact between two surfaces can exert two perpendicular components on the body you are modelling. The normal contact force, or normal reaction, acts perpendicular to the surface. Friction acts along the surface, opposing sliding or the tendency to slide relative to it. A smooth surface is modelled as frictionless.

[DIAGRAM: asset_name: Contact force and its two components; asset_slug: h645_b056_contact; caption: Normal reaction and friction are perpendicular components of one total contact force; description: A block on horizontal rough ground has upward normal reaction R, leftward friction F, rightward applied force P and downward weight W. A separate vector triangle adds leftward F and upward R to a diagonal total contact force C. It states C squared equals R squared plus F squared.]
Normal reaction and friction are perpendicular components of one total contact force

Total contact force

The total contact force is the vector sum of the normal reaction and the frictional force. If their magnitudes are RR and FF, its magnitude is

C=R2+F2C=\sqrt{R^2+F^2}

For example, R=12 NR=12\,\mathrm N, F=5 NF=5\,\mathrm N give C=13 NC=13\,\mathrm N, not 17 N17\,\mathrm N. With no friction, the total contact force is just the normal reaction.

The magnitude relation is Pythagoras' theorem and is not in the formula booklet: learn it. If the direction is required, define the reference line: the angle β\beta between the total contact force and the normal satisfies tan⁡β=F/R\tan\beta=F/R, for R>0R>0. The complementary angle is the angle to the plane; confusing those two reference lines gives the wrong direction.

The normal reaction need not equal the weight. On a horizontal surface, an upward component of a pull reduces RR; a downward push increases it. A non-adhesive surface can push but cannot pull, so R≥0R\ge0. A calculation giving R<0R<0 means the assumed contact cannot be maintained.

Key idea

Draw either the total contact force or its normal and frictional components on a force diagram. Drawing all three as separate forces would count the same contact twice.

A force diagram shows forces on the selected body only. Its equal-and-opposite forces on the surface belong to the surface's diagram, not to the block's.

Resting friction adjusts up to a limit

A rough surface does not automatically exert its maximum friction. A book resting on a horizontal table with no horizontal force has friction 00, even though the surfaces are rough. When a small horizontal push is applied, friction can adjust to balance it.

The friction model

For coefficient of friction μ≥0\mu\ge0 and normal reaction R≥0R\ge0, a resting contact can supply a frictional magnitude satisfying

0≤F≤μR0\le F\le\mu R

At limiting equilibrium, when sliding is just about to occur, F=μRF=\mu R. During sliding, this course models the frictional magnitude by F=μRF=\mu R, directed against the relative sliding velocity.

Not in the formula booklet: learn it.

The coefficient μ\mu is dimensionless. Its value is the limiting ratio F/RF/R in this model; there is no general restriction μ≤1\mu\le1. Treating it as constant and using the same simple relation throughout sliding are modelling assumptions, not claims that every real contact behaves identically at every speed or condition.

Worked example 1: friction below its maximum

Question. A particle of weight 50 N50\,\mathrm N rests on a rough horizontal surface with μ=0.4\mu=0.4. A horizontal force of 12 N12\,\mathrm N acts to the right. Determine whether equilibrium is possible and find the friction if it is.

Plan. Find the normal reaction, calculate the friction needed for equilibrium, then compare it with the maximum.

R=50vertical equilibriumFrequired=12horizontal balanceμR=0.4(50)=20maximum available12<20required friction is available\begin{aligned} R&=50 && \text{vertical equilibrium} \\ F_{\mathrm{required}}&=12 && \text{horizontal balance} \\ \mu R&=0.4(50)=20 && \text{maximum available} \\ 12&<20 && \text{required friction is available} \end{aligned}

Why this step? Equilibrium determines the actual friction; μR\mu R only checks whether the surface can supply it.

Answer. Equilibrium is possible, with friction 12 N12\,\mathrm N to the left. The contact is not limiting.

Check. The horizontal resultant is 12−12=012-12=0, and F/R=12/50=0.24<0.4F/R=12/50=0.24<0.4.

Change only the horizontal push, from 1212 to 24 N24\,\mathrm N. The normal reaction and friction limit stay the same.

Worked example 2: more friction is required than available

Question. For the same particle and surface, a horizontal force of 24 N24\,\mathrm N acts to the right. Determine whether it can remain at rest.

Plan. Compare the equilibrium requirement with the unchanged maximum friction.

R=50vertical balanceFrequired=24force needed to balance the pushμR=20same contact limit24>20equilibrium inequality fails\begin{aligned} R&=50 && \text{vertical balance} \\ F_{\mathrm{required}}&=24 && \text{force needed to balance the push} \\ \mu R&=20 && \text{same contact limit} \\ 24&>20 && \text{equilibrium inequality fails} \end{aligned}

Why this step? Writing F=24F=24 would describe a force the model cannot supply. Once sliding starts, its modelled magnitude is 2020, not 2424.

Answer. It cannot remain at rest; it begins sliding to the right.

Check. During rightward sliding, the horizontal resultant is 24−20=4 N24-20=4\,\mathrm N to the right, consistent with the predicted motion from rest.

At exactly 20 N20\,\mathrm N, the particle is in limiting equilibrium. Equality describes the threshold; strict inequality describes a reserve of available friction.

Watch out

Wrong: “The body is at rest, so F=μRF=\mu R.”

Right: resting friction satisfies F≤μRF\le\mu R. Use equality only at a specified or established limiting case, or during modelled sliding.

Check the required force first, before assigning a friction value.

The slope at the point of slipping

Consider a particle on a plane inclined at α\alpha to the horizontal, acted on only by its weight and the plane's contact force. Resolve parallel and perpendicular to the plane. Weight supplies downhill component mgsin⁡αmg\sin\alpha and inward normal component mgcos⁡αmg\cos\alpha.

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