MF12 - Motion on rough surfaces

MF12 - Motion on rough surfaces

Apply Newton's laws to sliding on rough surfaces, including a force that changes normal contact. Separate uphill motion, first rest and any reversal, then extend the force model to connected particles.

Sliding on a horizontal surface

Would the same angled force make a box accelerate more when you pull upwards or push downwards? Its forward component can be identical in both cases, yet the accelerations differ because the normal reaction changes.

Take a 4 kg4\,\mathrm{kg} box with μ=0.25\mu=0.25 and g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. A 20 N20\,\mathrm N force with horizontal component 16 N16\,\mathrm N and vertical component 12 N12\,\mathrm N gives reaction 28 N28\,\mathrm N when pulling up, but 52 N52\,\mathrm N when pushing down. The forward drive is unchanged; friction is not.

You already know

Resolve the forces and use resultant force =ma=ma. For constant acceleration, v=u+atv=u+at and s=ut+12at2s=ut+\tfrac12at^2 are in the formula sheet. Use them only while the same acceleration applies.

A rough-surface sliding model uses F=μRF=\mu R, opposite sliding. Here the stated coefficient also supplies the static bound at rest. Model the box as a particle, ignoring toppling.

An angled force during rightward sliding

For an upward force PP at acute angle α\alpha to horizontal,

R=mg−Psin⁡αPcos⁡α−μR=maa=Pm(cos⁡α+μsin⁡α)−μg\begin{aligned} R&=mg-P\sin\alpha\\ P\cos\alpha-\mu R&=ma\\ a&=\frac Pm(\cos\alpha+\mu\sin\alpha)-\mu g \end{aligned}

For a downward force, instead use R=mg+Psin⁡αR=mg+P\sin\alpha, giving a=(P/m)(cos⁡α−μsin⁡α)−μga=(P/m)(\cos\alpha-\mu\sin\alpha)-\mu g. Require maintained contact and rightward sliding throughout the interval.

Not in the formula sheet: learn it.

The acceleration expressions are derived by eliminating RR; using the two component equations is just as effective.

Worked example 1: an upward force while sliding

Question. A 4 kg4\,\mathrm{kg} particle slides right at 2 m s−12\,\mathrm{m\,s^{-1}} on rough horizontal ground with μ=0.25\mu=0.25. A 20 N20\,\mathrm N force acts right and up, with cosine 4/54/5 and sine 3/53/5 relative to horizontal. Take g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Determine acceleration and speed after 2 s2\,\mathrm s, assuming contact persists.

Plan. Find the reduced reaction and friction, then use the horizontal resultant and velocity equation.

R=40−20(35)=28 Nnormal balanceF=14(28)=7 N leftopposes sliding16−7=4arightward second lawa=94 m s−2solvev=2+94(2)=6.5 m s−1v=u+at\begin{aligned} R&=40-20\left(\frac35\right)=28\,\mathrm N &&\text{normal balance}\\ F&=\frac14(28)=7\,\mathrm N\text{ left} &&\text{opposes sliding}\\ 16-7&=4a &&\text{rightward second law}\\ a&=\frac94\,\mathrm{m\,s^{-2}} &&\text{solve}\\ v&=2+\frac94(2)=6.5\,\mathrm{m\,s^{-1}} &&v=u+at \end{aligned}

Why include the vertical component? It reduces friction even though it creates no vertical acceleration.

Answer. Acceleration is 2.25 m s−22.25\,\mathrm{m\,s^{-2}} right, and speed after 2 s2\,\mathrm s is 6.5 m s−16.5\,\mathrm{m\,s^{-1}}.

Check. R>0R>0, velocity remains rightwards, and ma=9 Nma=9\,\mathrm N, the horizontal resultant.

Change only the vertical sense of the applied force, so it pushes downwards at the same angle. Determine the same quantities after the same time.

Worked example 2: a downward force while sliding

Question. The same 4 kg4\,\mathrm{kg} particle starts at 2 m s−12\,\mathrm{m\,s^{-1}} right with μ=0.25\mu=0.25. The 20 N20\,\mathrm N force now acts right and down, with cosine 4/54/5 and sine 3/53/5. Take g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Determine acceleration and speed after 2 s2\,\mathrm s, assuming contact persists.

Plan. Increase the reaction by the downward component before calculating friction.

R=40+12=52 Nnormal balanceF=14(52)=13 N leftsliding friction16−13=4arightward second lawa=34 m s−2solvev=2+34(2)=3.5 m s−1v=u+at\begin{aligned} R&=40+12=52\,\mathrm N &&\text{normal balance}\\ F&=\frac14(52)=13\,\mathrm N\text{ left} &&\text{sliding friction}\\ 16-13&=4a &&\text{rightward second law}\\ a&=\frac34\,\mathrm{m\,s^{-2}} &&\text{solve}\\ v&=2+\frac34(2)=3.5\,\mathrm{m\,s^{-1}} &&v=u+at \end{aligned}

Why is the acceleration smaller? The same horizontal drive must now overcome 6 N6\,\mathrm N more friction.

Answer. Acceleration is 0.75 m s−20.75\,\mathrm{m\,s^{-2}} right, and speed after 2 s2\,\mathrm s is 3.5 m s−13.5\,\mathrm{m\,s^{-1}}.

Check. Contact and rightward sliding persist. The 3 N3\,\mathrm N resultant equals 4(0.75) N4(0.75)\,\mathrm N.

The opening question is settled: lifting reduces friction and gives greater acceleration in these two models. The force need not lift the particle completely off the ground to matter.

Watch out

Wrong: friction opposes acceleration. Right: friction opposes sliding. A particle moving right may accelerate left while its friction also points left.

Try a push strong enough vertically to make the rightward sliding slow down.

Motion down a rough incline

Sliding downhill does not necessarily mean speeding up downhill. Weight drives down the slope, friction opposes it, and their difference determines the acceleration.

Use the line of greatest slope for the motion here. Normal equilibrium gives R=mgcos⁡θR=mg\cos\theta, where θ\theta is the acute inclination to the horizontal. Friction acts uphill while sliding is downhill.

The downhill equation

With only weight and contact, and downhill positive,

mgsin⁡θ−μmgcos⁡θ=ma,a=g(sin⁡θ−μcos⁡θ)mg\sin\theta-\mu mg\cos\theta=ma, \qquad a=g(\sin\theta-\mu\cos\theta)

The mass cancels. This expression is valid only during downhill sliding; if the particle stops, test static equilibrium before continuing.

Not in the formula sheet: learn it.

Mass-independent acceleration is a useful consequence of this model. It relies on both weight and sliding friction being proportional to mass.

Worked example 3: gravity wins along the slope

Question. A 6 kg6\,\mathrm{kg} particle is projected downhill at 3 m s−13\,\mathrm{m\,s^{-1}} on a plane with sin⁡θ=5/13\sin\theta=5/13, cos⁡θ=12/13\cos\theta=12/13, and μ=1/4\mu=1/4. Only weight and contact act, with g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Determine its downhill acceleration and distance travelled in the first second. Give exact values.

Plan. Resolve weight and friction, then use the signed acceleration to find displacement.

R=72013 Nnormal balanceF=18013 N uphillsliding friction30013−18013=6adownhill second lawa=2013 m s−2solves=3+12(2013)=4913 mone-second displacement\begin{aligned} R&=\frac{720}{13}\,\mathrm N &&\text{normal balance}\\ F&=\frac{180}{13}\,\mathrm N\text{ uphill} &&\text{sliding friction}\\ \frac{300}{13}-\frac{180}{13}&=6a &&\text{downhill second law}\\ a&=\frac{20}{13}\,\mathrm{m\,s^{-2}} &&\text{solve}\\ s&=3+\frac12\left(\frac{20}{13}\right)=\frac{49}{13}\,\mathrm m &&\text{one-second displacement} \end{aligned}

Why does displacement equal distance? The velocity stays downhill throughout the interval.

Answer. Downhill acceleration is 20/13 m s−220/13\,\mathrm{m\,s^{-2}}, and distance is 49/13 m49/13\,\mathrm m.

Check. Final velocity is 3+20/13=59/13 m s−13+20/13=59/13\,\mathrm{m\,s^{-1}} downhill, giving average speed 49/13 m s−149/13\,\mathrm{m\,s^{-1}}.

Change only μ\mu, from 1/41/4 to 1/21/2. Determine the same quantities over the same second.

Worked example 4: friction wins while motion stays downhill

Question. The same 6 kg6\,\mathrm{kg} particle is projected downhill at 3 m s−13\,\mathrm{m\,s^{-1}} on the same plane, but μ=1/2\mu=1/2. Only weight and contact act, with g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Determine its downhill acceleration and distance travelled in the first second. Give exact values.

Plan. Recompute uphill friction and check that downhill velocity stays positive.

R=72013 Nnormal balanceF=36013 N uphilllarger friction30013−36013=6adownhill second lawa=−1013 m s−2solves=3−12(1013)=3413 mone-second displacement\begin{aligned} R&=\frac{720}{13}\,\mathrm N &&\text{normal balance}\\ F&=\frac{360}{13}\,\mathrm N\text{ uphill} &&\text{larger friction}\\ \frac{300}{13}-\frac{360}{13}&=6a &&\text{downhill second law}\\ a&=-\frac{10}{13}\,\mathrm{m\,s^{-2}} &&\text{solve}\\ s&=3-\frac12\left(\frac{10}{13}\right)=\frac{34}{13}\,\mathrm m &&\text{one-second displacement} \end{aligned}

Why keep friction uphill? The negative acceleration means slowing down, not that the particle has already reversed.

Answer. Downhill acceleration is −10/13 m s−2-10/13\,\mathrm{m\,s^{-2}}, meaning 10/13 m s−210/13\,\mathrm{m\,s^{-2}} uphill. Distance is 34/13 m34/13\,\mathrm m.

Check. Final velocity is 29/13 m s−129/13\,\mathrm{m\,s^{-1}} downhill. First rest is at 39/10 s39/10\,\mathrm s, after the interval used.

The coefficient changes the acceleration without changing the sliding-friction direction. That direction changes only if sliding reverses.

Watch out

Wrong: substitute the weight mgmg into F=μRF=\mu R on an incline. Right: first find the normal reaction mgcos⁡θmg\cos\theta, then calculate friction.

Try a slope given by its angle rather than a component ratio.

Stopping, holding or reversing on an incline

An uphill launch has two phases to investigate: sliding until first rest, then whatever the static force comparison allows. Reaching zero velocity does not guarantee a reversal.

At the boundary μ=tan⁡θ\mu=\tan\theta, friction can just hold the particle after it stops. This is limiting equilibrium, not a command to move downhill. The same model predicts no spontaneous motion from a balanced state.

[DIAGRAM: asset_name: Friction changes at reversal; asset_slug: incline_friction_stages; caption: First rest separates the uphill equation from the static decision and any downhill equation; description: Three panels on the same incline show uphill sliding with friction downhill, first rest with required uphill friction tested against capacity, and downhill sliding with friction uphill if capacity was insufficient.]
First rest separates the uphill equation from the static decision and any downhill equation

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