MF10, MF11 - Contact forces, friction and limiting equilibrium

MF10, MF11 - Contact forces, friction and limiting equilibrium

Contact with a rough surface supplies a normal push and friction along the surface. You will interpret their combined vector, check whether a proposed contact force is physically possible, and use different friction rules at rest, at the limit and during sliding.

Normal force, friction and overall contact

Can friction point in the same direction as acceleration? Push a sliding particle right while it is moving left: both your push and friction can point right, slowing the leftward motion. To make sense of this, separate motion, resultant force and the individual contact components.

You already know

A vector has signed components but a non-negative magnitude. Perpendicular components combine by Pythagoras. A force diagram contains forces acting on the selected body, with arrows showing their directions.

A rough surface's contact is one interaction described by two useful components.

Normal and frictional contact

The normal contact force is perpendicular to the surface and pushes away from it, so its magnitude RR cannot be negative. The frictional force is tangential and opposes sliding or its tendency. Smooth means friction may be ignored; rough means friction must be taken into account. Rough does not mean that friction must have its maximum value.

For a horizontal surface, with i\mathbf i right and j\mathbf j up, the vector −12i+35j N-12\mathbf i+35\mathbf j\,\mathrm N consists of 12 N12\,\mathrm N friction left and 35 N35\,\mathrm N normal force up. Its overall magnitude is 37 N37\,\mathrm N.

[DIAGRAM: asset_name: Components of rough contact; asset_slug: rough_contact_components; description: Equal force-component scales show friction (-12,0), normal reaction (0,35) and their resultant (-12,35) newtons, clearly marked as alternative descriptions; caption: Friction and normal reaction combine to give the overall contact force]
Friction and normal reaction combine to give the overall contact force

Use either the overall vector or its two components in a force sum. Including all three would count the interaction twice. The contact decomposition and the vector-magnitude rule are not in the formula sheet.

Worked example 1: checking a proposed contact force

Question. A calculation proposes contact force C=−9i+12j N\mathbf C=-9\mathbf i+12\mathbf j\,\mathrm N on a particle above horizontal ground, with i\mathbf i right and j\mathbf j up. Identify the normal and friction components, calculate the vector's magnitude, and state whether its normal component is compatible with pushing contact.

Plan. Separate the parallel and perpendicular components, then check the sign of the normal component.

R=12upward normal componentFfriction=−9itangential component∣C∣=92+122=15component triangleR>0pushing contact is possible\begin{aligned} R&=12 &&\text{upward normal component} \\ \mathbf F_{\text{friction}}&=-9\mathbf i &&\text{tangential component} \\ |\mathbf C|&=\sqrt{9^2+12^2}=15 &&\text{component triangle} \\ R&>0 &&\text{pushing contact is possible} \end{aligned}

Why check the sign? An algebraically defined vector need not represent a physically possible force from an unglued surface.

Answer. The normal force is 12 N12\,\mathrm N upwards; friction is 9 N9\,\mathrm N left; overall magnitude is 15 N15\,\mathrm N. Its normal component is compatible with pushing contact. A friction-capacity check would require further information.

Check. The normal component points away from the ground, as a supporting push should.

Change only the sign of the proposed normal component. The vector still has a magnitude, but its physical interpretation changes.

Worked example 2: an impossible normal component

Question. The calculation instead proposes C=−9i−12j N\mathbf C=-9\mathbf i-12\mathbf j\,\mathrm N. Identify the normal and friction components, calculate the vector's magnitude, and state whether its normal component is compatible with pushing contact.

Plan. Perform the same component calculation and test the perpendicular sign.

Cy=−12points into the groundCparallel=−9ileftward tangential component∣C∣=92+(−12)2=15vector magnitude still existsCy<0not a supporting normal push\begin{aligned} C_y&=-12 &&\text{points into the ground} \\ \mathbf C_{\text{parallel}}&=-9\mathbf i &&\text{leftward tangential component} \\ |\mathbf C|&=\sqrt{9^2+(-12)^2}=15 &&\text{vector magnitude still exists} \\ C_y&<0 &&\text{not a supporting normal push} \end{aligned}

Why reject the model? The surface cannot supply a downward pulling normal force on a particle above it.

Answer. The proposed normal component is −12 N-12\,\mathrm N, the tangential component is 9 N9\,\mathrm N left, and the proposed vector has magnitude 15 N15\,\mathrm N. It cannot be the contact force under this pushing-contact model.

Check. Taking an absolute value would hide the physical failure rather than repair it.

When the normal reaction reaches zero, the surface provides no supporting push and the contact assumption needs checking. A computed negative value cannot be repaired by simply drawing the normal arrow downwards.

Watch out

Wrong: replace a calculated negative reaction by its positive magnitude and continue. Correct: reject or revise the assumed contact, since a normal push cannot pull the particle onto the surface.

Now interpret a physically possible contact vector.

Actual friction and its upper bound

For a particle to remain at rest, contact must supply the required normal push and enough friction to balance tangential forces.

The coefficient of friction μ\mu is dimensionless. It specifies a ratio in an ideal model, not the value of friction by itself.

Static friction and its limit

While contact is maintained, static friction magnitude FF satisfies

0≤F≤μR0\le F\le\mu R

The limiting value is μR\mu R. Limiting equilibrium is equilibrium on the point of sliding, with F=μRF=\mu R. Friction acts opposite the tendency to slide.

Not in the formula sheet: learn it.

Calculate the friction required for rest first. Then compare with the capacity; setting equality at the outset would assume limiting equilibrium.

Worked example 3: unused friction capacity

Question. A 5 kg5\,\mathrm{kg} box rests on horizontal ground with coefficient μ=0.4\mu=0.4. A 12 N12\,\mathrm N horizontal force acts right, and the only other forces are weight and contact. Use g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Find normal reaction, friction required, limiting friction, and whether rest is possible and limiting.

Plan. Balance perpendicular and parallel to the ground, then test the inequality.

R=50vertical balanceFrequired=12horizontal balanceμR=0.4(50)=20limiting capacity12<20required friction is within the bound\begin{aligned} R&=50 &&\text{vertical balance} \\ F_{\text{required}}&=12 &&\text{horizontal balance} \\ \mu R&=0.4(50)=20 &&\text{limiting capacity} \\ 12&<20 &&\text{required friction is within the bound} \end{aligned}

Why distinguish the two values? Actual friction is determined by balance, while limiting friction is a capacity.

Answer. R=50 NR=50\,\mathrm N; friction is 12 N12\,\mathrm N left; limiting friction is 20 N20\,\mathrm N. Rest is possible and not limiting.

Check. The horizontal resultant is 12−12=012-12=0, rather than 12−2012-20.

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