4.1 - Friction and limiting equilibrium

4.1 - Friction and limiting equilibrium

A gentle pull on a rough surface may produce no motion at all: friction can adjust to the force needed for balance. You will determine its direction and magnitude, check whether rest is possible, and use limiting equilibrium when a particle is about to slip. Resolving perpendicular to the surface matters just as much as resolving along it.

Normal and frictional contact

A box rests on a rough slope, held by an uphill pull. Does friction act uphill or downhill? It depends on which way the box would tend to slide without friction, rather than simply on the direction of the pull.

The contact force has a normal component RR, perpendicular to the plane and away from it, and a frictional component FF, along the plane. For a body at rest, friction opposes the tendency for relative sliding. For a body sliding along a fixed plane, it opposes that sliding. Friction can be zero if no tangential force is needed.

Frictional contact

Friction is the tangential component of contact between surfaces. The normal contact force is the perpendicular component. Together they form the total contact force; on a rough surface that total need not be perpendicular to the plane.

Take axes along and perpendicular to the plane. Weight mgmg contributes mgsin⁡αmg\sin\alpha downhill and mgcos⁡αmg\cos\alpha into a plane inclined at α\alpha to the horizontal.

[DIAGRAM: asset_name: Contact and weight on a slope; asset_slug: mechanics-friction-plane; caption: Weight resolves into a downhill component and an inward normal component. friction depends on the tendency to slide; description: Exact geometry for a 30 degree slope shows a particle with weight 100 N vertically down, normal contact perpendicular away from the slope, a 20 N pull uphill along the slope and friction uphill. A separate component triangle resolves weight into 50 N downhill and 50 square root 3 N into the plane. The force diagram contains weight itself rather than also adding its components as extra forces.]
Weight resolves into a downhill component and an inward normal component. friction depends on the tendency to slide

Worked example 1: friction acts uphill

Question. A particle of mass 10 kg10\,\mathrm{kg} rests on a rough plane inclined at 30∘30^\circ to the horizontal. A force of 20 N20\,\mathrm{N} pulls it directly up the plane. Find RR and the friction needed for rest.

Plan. Resolve perpendicular to the plane, then compare the pull with the downhill weight component.

R=100cos⁡30∘=503 Ndownhill weight component=100sin⁡30∘=50 NF+20=50friction must act uphillF=30 N\begin{aligned} R&=100\cos30^\circ=50\sqrt3\,\mathrm{N}\\ \text{downhill weight component}&=100\sin30^\circ=50\,\mathrm{N}\\ F+20&=50 &&\text{friction must act uphill}\\ F&=30\,\mathrm{N} \end{aligned}

Answer. R=503 NR=50\sqrt3\,\mathrm{N}; friction 30 N30\,\mathrm{N} uphill is required.

Check. The uphill components total 30+20=50 N30+20=50\,\mathrm{N}, balancing the downhill component. Whether the surface can supply this friction is a separate question.

Change only the pull to 60 N60\,\mathrm{N}. It now exceeds the downhill weight component, so the required friction reverses.

Worked example 2: friction acts downhill

Question. The same particle rests on the same plane, now pulled directly uphill by 60 N60\,\mathrm{N}. Find RR and the friction needed for rest.

Plan. The perpendicular balance is unchanged; use downhill friction to balance the excess pull.

R=503 N60=50+Fbalance along the planeF=10 N\begin{aligned} R&=50\sqrt3\,\mathrm{N}\\ 60&=50+F &&\text{balance along the plane}\\ F&=10\,\mathrm{N} \end{aligned}

Answer. R=503 NR=50\sqrt3\,\mathrm{N}; friction 10 N10\,\mathrm{N} downhill is required.

Check. Omitting friction would leave 60−50=10 N60-50=10\,\mathrm{N} uphill. Downhill friction opposes precisely that tendency.

If you assume a direction and obtain a negative signed friction, reverse your assumed arrow and report the positive magnitude. Do not discard a valid calculation just because the initial guess was wrong.

Watch out

Do not draw weight and both of its components as three separate forces on the same diagram. The components replace weight in the equations; they are not extra interactions.

Compare the uphill pull with the weight component in a fresh situation.

Friction below the limit

For a particle at rest, first calculate the friction required by equilibrium. Then check whether the surface can provide it. The maximum available friction is not automatically the actual friction.

Coefficient of friction

The coefficient of friction μ\mu is the ratio of the limiting friction to the normal contact force, for R>0R>0. It has no units. In this model it is a property of the pair of contacting surfaces.

For example, if R=50 NR=50\,\mathrm{N} and μ=0.4\mu=0.4, the limiting friction is 20 N20\,\mathrm{N}. A required friction of 12 N12\,\mathrm{N} is possible; a required friction of 24 N24\,\mathrm{N} is not.

Static friction bound

For a particle remaining at rest in rough contact,

0≤F≤μR,0\le F\le\mu R,

where FF is the magnitude of friction. Below the limit, use balance to find FF. At the limit, F=μRF=\mu R.

Not in the formula booklet: learn it.

An impossible equilibrium tells you that the particle will begin to slip under the model. The bound alone does not calculate its acceleration.

Worked example 3: enough friction is available

Question. A 5 kg5\,\mathrm{kg} particle is on a rough horizontal plane with μ=0.4\mu=0.4. A horizontal force of 12 N12\,\mathrm{N} pulls it rightwards. Determine whether it can remain at rest and find the friction if it does.

Plan. Find the normal force, calculate the friction bound and compare with the required force.

R=5×10=50 NFmax⁡=0.4×50=20 NFrequired=12 Nhorizontal balance\begin{aligned} R&=5\times10=50\,\mathrm{N}\\ F_{\max}&=0.4\times50=20\,\mathrm{N}\\ F_{\text{required}}&=12\,\mathrm{N} &&\text{horizontal balance} \end{aligned}

Answer. Rest is possible because 12<2012<20. Friction is 12 N12\,\mathrm{N} leftwards.

Check. Horizontal resultant is 12−12=012-12=0. Using F=20F=20 would instead give an unbalanced leftward force and contradict rest.

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