R6 - Friction and limiting equilibrium

R6 - Friction and limiting equilibrium

Find the friction actually needed for rest, compare it with the available capacity and recognise limiting equilibrium. You will resolve angled forces, check contact, and find complete ranges of coefficients or tensions on rough horizontal and inclined surfaces.

Friction direction and capacity

Can the same rough surface exert friction to the left, to the right or not at all? Put a block on a level table and pull it gently right. If it stays still, friction balances your pull to the left. Pull left instead and friction acts right. With no horizontal pull and no other horizontal force, no friction is needed.

You already know

Equilibrium means zero resultant in each direction. A normal reaction acts perpendicular to a surface; resolving separates parallel and perpendicular components. For a force PP at angle θ\theta above horizontal, those components are Pcos⁡θP\cos\theta horizontally and Psin⁡θP\sin\theta vertically. These relationships are not in the formula booklet: learn them.

Friction acts parallel to the contact surface and opposes relative sliding or the tendency to slide. For a resting block, find the force needed for balance before comparing it with the surface's capacity.

Static friction model

0≤F≤μR0\le F\le\mu R

Here FF is the magnitude of friction, R≥0R\ge0 is the normal reaction and μ≥0\mu\ge0 is the coefficient of friction. The product μR\mu R is the maximum available friction in this model. Not in the formula booklet: learn it.

The coefficient has no units: at the limiting boundary it is the ratio of two force magnitudes. A rough surface does not automatically exert its maximum friction, just as a support does not automatically carry its greatest possible load.

Worked example 1: actual friction below capacity

Question. A block of weight 40 N40\,\mathrm N rests on a rough horizontal surface. A horizontal 12 N12\,\mathrm N force pulls right. The coefficient of friction is 0.400.40, and no other forces act. Find the actual friction, its maximum available value and whether equilibrium is limiting.

Plan. Balance perpendicular forces to obtain the reaction, balance horizontal forces to obtain actual friction, then compare demand with capacity.

R=40 Nvertical balanceF=12 N leftbalance the rightward pullμR=0.40(40)=16 Nmaximum available12<16demand is below capacity\begin{aligned} R&=40\,\mathrm N && \text{vertical balance} \\ F&=12\,\mathrm N\text{ left} && \text{balance the rightward pull} \\ \mu R&=0.40(40)=16\,\mathrm N && \text{maximum available} \\ 12&<16 && \text{demand is below capacity} \end{aligned}

Why this step? Equilibrium determines the actual friction. The coefficient only checks whether that required force can be supplied.

Answer. Friction is 12 N12\,\mathrm N left, its capacity is 16 N16\,\mathrm N, and equilibrium is not limiting.

Check. The horizontal resultant is 12−12=012-12=0. Using 16 N16\,\mathrm N as the actual friction would create an unwanted leftward resultant.

Worked example 2 reverses only the pull. The reaction and capacity stay unchanged; the friction direction must reverse.

Worked example 2: the same surface, opposite friction

Question. The same 40 N40\,\mathrm N block rests on the same surface with μ=0.40\mu=0.40, now under a horizontal 12 N12\,\mathrm N pull to the left. No other forces act. Find the friction and decide whether rest is possible.

Plan. Balance the new pull with friction in the opposite direction and check its magnitude against the unchanged capacity.

R=40 Nvertical forces are unchangedF=12 N righthorizontal balanceF=12<0.40(40)=16capacity is sufficient\begin{aligned} R&=40\,\mathrm N && \text{vertical forces are unchanged} \\ F&=12\,\mathrm N\text{ right} && \text{horizontal balance} \\ F&=12<0.40(40)=16 && \text{capacity is sufficient} \end{aligned}

Why this step? The tendency to slide has reversed. “Rough horizontal surface” does not specify a fixed leftward friction arrow.

Answer. Friction is 12 N12\,\mathrm N right; rest is possible below the limiting value.

Check. Taking rightwards positive, −12+12=0-12+12=0. If the pull were removed, equilibrium would need F=0F=0, also allowed by the inequality.

The model can admit rest at the maximum too. Equality is a special boundary condition, explored next; it is not a consequence of being stationary.

Watch out

Wrong: F=0.40(40)=16 NF=0.40(40)=16\,\mathrm N simply because the block rests on that surface. Right: equilibrium needs F=12 NF=12\,\mathrm N; then check 12≤1612\le16. Setting capacity equal to actual friction without a limiting condition changes the force balance.

In the next case, calculate both quantities and explain their different roles.

Limiting equilibrium and coefficients

A block is in limiting equilibrium when it remains at rest but is on the verge of slipping. Friction has then reached its maximum available magnitude. The impending direction determines which way that limiting friction acts.

Limiting condition and coefficient bound

At limiting equilibrium,

F=μRF=\mu R

For equilibrium with known required friction FF and reaction R>0R>0,

μ≥FR\mu\ge\frac{F}{R}

The least possible coefficient is the included boundary F/RF/R. Not in the formula booklet: learn it.

The inequality follows by dividing F≤μRF\le\mu R by positive RR. A larger coefficient increases available capacity; it does not force actual static friction to increase when the other forces are unchanged.

Worked example 3: a limiting measurement

Question. A block of weight 75 N75\,\mathrm N is on a rough horizontal surface. It is just about to slide under a horizontal 21 N21\,\mathrm N pull. Weight, reaction, pull and friction are its only forces. Find the coefficient of friction.

Plan. Use equilibrium to find reaction and required friction, then use the stated limiting condition.

R=75 Nvertical balanceF=21 Nhorizontal balance21=75μlimiting frictionμ=0.28\begin{aligned} R&=75\,\mathrm N && \text{vertical balance} \\ F&=21\,\mathrm N && \text{horizontal balance} \\ 21&=75\mu && \text{limiting friction} \\ \mu&=0.28 \end{aligned}

Why this step? “Just about to slide” supplies the equality. Rest without that extra information would give a lower bound instead.

Answer. The coefficient is 0.280.28.

Check. 0.28(75)=21 N0.28(75)=21\,\mathrm N, exactly the needed friction.

Worked example 4 changes the information about equilibrium: only rest under a smaller pull is reported, and the surface's coefficient is now unknown. Find a range, rather than declaring one value.

Worked example 4: all coefficients that allow rest

Question. A block of weight 75 N75\,\mathrm N rests on a rough horizontal surface under a horizontal 15 N15\,\mathrm N pull. The coefficient μ\mu is unknown and no limiting condition is stated. No other forces act. Find the full range of coefficients compatible with rest.

Plan. Find the actual friction, then require enough capacity to supply it.

R=75 N,F=15 Nforce balance15≤75μrequired friction must be availableμ≥0.20divide by positive 75\begin{aligned} R&=75\,\mathrm N,\qquad F=15\,\mathrm N && \text{force balance} \\ 15&\le75\mu && \text{required friction must be available} \\ \mu&\ge0.20 && \text{divide by positive }75 \end{aligned}

Why this step? Equality gives the smallest permitted coefficient; coefficients larger than that still supply the same actual 15 N15\,\mathrm N friction.

Answer. μ≥0.20\mu\ge0.20.

Check. At μ=0.20\mu=0.20, capacity is 15 N15\,\mathrm N and equilibrium is limiting. At μ=0.28\mu=0.28, capacity is 21 N21\,\mathrm N, so rest remains possible without reaching the limit.

“Least possible” includes the limiting case. A strict inequality would remove the very minimum you were asked to find.

Watch out

Wrong: “The block is at rest, so μ=F/R\mu=F/R.” Right: rest requires μ≥F/R\mu\ge F/R for R>0R>0. Equality requires limiting equilibrium or identifies the least compatible coefficient; it is not forced by rest alone.

When R=0R=0, do not divide by it. The friction model gives zero capacity, so equilibrium then requires zero tangential demand as well. A negative reaction would mean the assumed contact cannot hold.

Angled forces change friction capacity

Pulling at an angle affects two things at once. Its horizontal component creates the friction demand, while its vertical component changes the normal reaction and therefore the capacity. Resolve first; do not set the reaction equal to weight automatically.

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