3.03v - Motion on rough surfaces

3.03v - Motion on rough surfaces

Use friction in directed equations of motion, with its direction set by sliding. Find accelerations on level ground and inclines, reconsider the forces when motion stops, and solve connected-particle problems with rough contact.

Sliding on a horizontal surface

A crate slides right, slows down and stops. Does its leftward friction then make it accelerate backwards? It does not. The equation for rightward sliding ends when the crate stops; static friction must then be reconsidered.

For a 5 kg5\,\mathrm{kg} crate with μ=0.2\mu=0.2 and g=10 m s−2g=10\,\mathrm{m\,s^{-2}}, friction during sliding is 10 N10\,\mathrm N. Without a drive, an initial speed of 6 m s−16\,\mathrm{m\,s^{-1}} falls to zero after 3 s3\,\mathrm s. After that, zero horizontal applied force needs zero static friction, so the crate can stay still.

You already know

Choose a positive direction and use resultant force =ma=ma. Constant acceleration allows v=u+atv=u+at and s=ut+12at2s=ut+\tfrac12at^2; these kinematic equations are in the formula sheet. They apply only while the stated acceleration persists.

Throughout this lesson, sliding friction is modelled as F=μRF=\mu R, with the same stated coefficient used in the static bound. Real surfaces can need a more detailed model.

Sliding on rough horizontal ground

For rightward sliding under a horizontal rightward drive PP, with no other vertical force,

R=mg,F=μmg leftwards,P−μmg=maR=mg, \qquad F=\mu mg\text{ leftwards}, \qquad P-\mu mg=ma

The friction direction follows sliding velocity, even when the acceleration is negative. Reassess the model at first rest.

Not in the formula sheet: learn it.

Begin with a drive stronger than friction.

Worked example 1: speeding up while sliding

Question. A 5 kg5\,\mathrm{kg} particle slides right at 6 m s−16\,\mathrm{m\,s^{-1}} on rough horizontal ground. A horizontal drive is 20 N20\,\mathrm N right, μ=0.2\mu=0.2, and g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Find its acceleration and its speed after 2 s2\,\mathrm s, assuming the conditions persist.

Plan. Calculate friction, apply Newton's second law rightwards, then use the velocity equation.

R=50 Nvertical balanceF=0.2(50)=10 N leftopposes sliding20−10=5arightward second lawa=2 m s−2solvev=6+2(2)=10 m s−1v=u+at\begin{aligned} R&=50\,\mathrm N &&\text{vertical balance}\\ F&=0.2(50)=10\,\mathrm N\text{ left} &&\text{opposes sliding}\\ 20-10&=5a &&\text{rightward second law}\\ a&=2\,\mathrm{m\,s^{-2}} &&\text{solve}\\ v&=6+2(2)=10\,\mathrm{m\,s^{-1}} &&v=u+at \end{aligned}

Why subtract friction? It opposes the rightward velocity, not the direction in which you hope the particle accelerates.

Answer. Acceleration is 2 m s−22\,\mathrm{m\,s^{-2}} right; the speed after 2 s2\,\mathrm s is 10 m s−110\,\mathrm{m\,s^{-1}}.

Check. The speed remains positive and increases, consistent with the positive 10 N10\,\mathrm N resultant.

Change only the drive, from 20 N20\,\mathrm N to zero. Find the same quantities after the same time.

Worked example 2: slowing down while still sliding right

Question. The same 5 kg5\,\mathrm{kg} particle starts at 6 m s−16\,\mathrm{m\,s^{-1}} right, with μ=0.2\mu=0.2 and g=10 m s−2g=10\,\mathrm{m\,s^{-2}}, but no drive acts. Find its acceleration and speed after 2 s2\,\mathrm s.

Plan. Use the unchanged friction as the whole horizontal resultant and check that the particle has not stopped.

R=50 Nvertical balanceF=10 N leftopposes rightward sliding−10=5arightward second lawa=−2 m s−2solvev=6−2(2)=2 m s−1v=u+at\begin{aligned} R&=50\,\mathrm N &&\text{vertical balance}\\ F&=10\,\mathrm N\text{ left} &&\text{opposes rightward sliding}\\ -10&=5a &&\text{rightward second law}\\ a&=-2\,\mathrm{m\,s^{-2}} &&\text{solve}\\ v&=6-2(2)=2\,\mathrm{m\,s^{-1}} &&v=u+at \end{aligned}

Why check the sign of velocity? A negative result would mean the assumed sliding stage ended before the requested time.

Answer. Acceleration is 2 m s−22\,\mathrm{m\,s^{-2}} left; the speed after 2 s2\,\mathrm s is 2 m s−12\,\mathrm{m\,s^{-1}}, still rightwards.

Check. First rest occurs at 6/2=3 s>2 s6/2=3\,\mathrm s>2\,\mathrm s. The distance to first rest is 6(3)−12(2)(32)=9 m6(3)-\tfrac12(2)(3^2)=9\,\mathrm m.

Beyond 3 s3\,\mathrm s, the second example stays at rest. Substituting a later time into its sliding equation would create a false reversal.

Watch out

Wrong: “Acceleration is leftwards, so friction must be rightwards.” Right: friction opposes sliding. A leftward resultant can slow a rightward-moving particle.

Try a smaller drive that still leaves a positive rightward speed.

Motion down a rough incline

A particle sliding downhill has uphill friction. It can speed up or slow down, depending on whether the downhill weight component exceeds friction.

We consider motion along the line of greatest slope. Resolve perpendicular to the plane to find the reaction, then parallel to it for the equation of motion.

Downhill sliding under weight and contact

For a plane at acute angle θ\theta, with downhill positive,

R=mgcos⁡θmgsin⁡θ−μmgcos⁡θ=maa=g(sin⁡θ−μcos⁡θ)\begin{aligned} R&=mg\cos\theta\\ mg\sin\theta-\mu mg\cos\theta&=ma\\ a&=g(\sin\theta-\mu\cos\theta) \end{aligned}

This holds while the particle slides downhill and remains in contact. Positive aa means speeding up downhill; negative aa means slowing down downhill.

Not in the formula sheet: learn it.

The acceleration formula comes from force balance; there is no need to memorise it instead of the diagram and equations.

Worked example 3: downhill motion speeding up

Question. A 3 kg3\,\mathrm{kg} particle is projected down a rough incline at 4 m s−14\,\mathrm{m\,s^{-1}}. Here sin⁡θ=3/5\sin\theta=3/5, cos⁡θ=4/5\cos\theta=4/5, μ=0.5\mu=0.5, and g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Find its downhill acceleration and the distance travelled in the first second.

Plan. Subtract uphill friction from downhill weight, then use constant-acceleration displacement.

R=30(45)=24 Nnormal balanceF=0.5(24)=12 N uphillsliding friction18−12=3adownhill second lawa=2 m s−2solves=4(1)+12(2)(12)=5 mconstant acceleration\begin{aligned} R&=30\left(\frac45\right)=24\,\mathrm N &&\text{normal balance}\\ F&=0.5(24)=12\,\mathrm N\text{ uphill} &&\text{sliding friction}\\ 18-12&=3a &&\text{downhill second law}\\ a&=2\,\mathrm{m\,s^{-2}} &&\text{solve}\\ s&=4(1)+\frac12(2)(1^2)=5\,\mathrm m &&\text{constant acceleration} \end{aligned}

Why resolve perpendicular first? Friction depends on RR, which is mgcos⁡θmg\cos\theta, not the whole weight.

Answer. The downhill acceleration is 2 m s−22\,\mathrm{m\,s^{-2}}, and the distance is 5 m5\,\mathrm m.

Check. Final velocity is 4+2(1)=6 m s−14+2(1)=6\,\mathrm{m\,s^{-1}} downhill. Average speed 5 m s−15\,\mathrm{m\,s^{-1}} gives the same distance.

Change only μ\mu, from 0.50.5 to 11. Keep downhill as positive and find the same outputs over one second.

Worked example 4: downhill motion slowing down

Question. The same 3 kg3\,\mathrm{kg} particle is projected downhill at 4 m s−14\,\mathrm{m\,s^{-1}} on the same plane, but μ=1\mu=1. Take g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Find its downhill acceleration and the distance travelled in the first second.

Plan. Recompute friction, retain its uphill direction, and check that downhill sliding lasts for the full second.

R=24 Nunchanged normal balanceF=24 N uphillnew friction18−24=3adownhill second lawa=−2 m s−2negative downhill accelerations=4(1)−12(2)(12)=3 mdisplacement while sliding\begin{aligned} R&=24\,\mathrm N &&\text{unchanged normal balance}\\ F&=24\,\mathrm N\text{ uphill} &&\text{new friction}\\ 18-24&=3a &&\text{downhill second law}\\ a&=-2\,\mathrm{m\,s^{-2}} &&\text{negative downhill acceleration}\\ s&=4(1)-\frac12(2)(1^2)=3\,\mathrm m &&\text{displacement while sliding} \end{aligned}

Why keep friction uphill? The particle still slides downhill, even though it is slowing.

Answer. The downhill acceleration is −2 m s−2-2\,\mathrm{m\,s^{-2}}, meaning 2 m s−22\,\mathrm{m\,s^{-2}} uphill. The distance is 3 m3\,\mathrm m.

Check. Final velocity is 2 m s−12\,\mathrm{m\,s^{-1}} downhill. First rest is at 2 s2\,\mathrm s, beyond the interval used.

For this course, incline-motion problems follow the line of greatest slope; resolving motion across the plane is unnecessary. The force diagram still needs both parallel and normal directions.

Watch out

Wrong: switch friction downhill when the calculated downhill acceleration is negative. Right: keep uphill friction until downhill sliding stops. Then reassess the static bound.

Try a different slope, using the same force-and-kinematics chain.

Stopping, holding or reversing on an incline

During uphill sliding, both friction and the downhill weight component oppose the motion. At the first stop, ask a new question: can static friction now balance the downhill pull?

If it can, the particle stays at rest. If it cannot, the particle begins sliding downhill and friction reverses. This can make the return journey slower than the climb, although gravity has not changed.

[DIAGRAM: asset_name: Friction changes at reversal; asset_slug: incline_friction_stages; description: Same rough incline shown in uphill sliding, first rest and downhill sliding stages, with separate force arrows, weight components and friction directions; caption: Reassess static friction at rest before choosing the next sliding equation]
Reassess static friction at rest before choosing the next sliding equation

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