Limiting equilibrium & coefficient of friction | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Use equality only at the limit

Definition

What is meant by limiting equilibrium?

Model answer: Equilibrium in which friction has reached its limiting value and the body is on the point of moving.

0\le F\le \mu R,\qquad F_{\max}=\mu R

Symbols

$F$ = actual friction while at rest (N)
$F_{\max}$ = limiting friction (N)
$\mu$ = coefficient of friction (none)
$R$ = normal reaction (N)

The coefficient of friction is defined by $\mu=F_{\max}/R$ . It measures the limiting friction per unit normal reaction in this model. A larger $\mu$ means more grip for the same $R$ ; $\mu$ has no unit.

At rest but not limiting: $F<\mu R$ .
On the point of moving: $F=\mu R$ .
While sliding in these Paper 4 models, use $F=\mu R$ opposite the motion.

Compare required friction with available friction

Swipe left or right to see the whole diagram →

For a body at rest, compare the friction required for equilibrium with the maximum friction available. The actual friction is the required value, not automatically μR.

Worked example

Can the block remain at rest?

A 5 kg block rests on a plane inclined at 20° to the horizontal. The coefficient of friction is 0.4 and no other force acts.

R=50\cos20^\circ=46.98\text{ N}

F_{\max}=0.4R=18.79\text{ N}

F_{\text{needed}}=50\sin20^\circ

F_{\text{needed}}=17.10\text{ N}

The required 17.10 N is below the available 18.79 N, so the block can remain at rest. Its actual friction is 17.10 N up the plane.

Answer

F=17.1\text{ N}

The friction acts up the plane, and equilibrium is possible.

Key idea

The test for rest is “friction needed ≤ friction available”. Do not replace the needed friction by

\mu R

before checking.

Recalculate R when the applied force is angled

An upward component of a pull reduces $R$ and therefore reduces limiting friction. A downward component increases both.

Worked example

A pull changes the normal reaction

A 10 kg block is on a horizontal floor with $\mu=0.25$ . It is about to move right under a force $P$ acting 30° above the horizontal.

R=100-P\sin30^\circ

P\cos30^\circ=0.25R

P\cos30^\circ=25-0.125P

Answer

P=25.2\text{ N}\quad\text{(3 s.f.)}

Your turnIndependent check [4 marks]

A block is in limiting equilibrium on a plane inclined at 30° to the horizontal, with no force other than weight and contact from the plane. Find the coefficient of friction.

Show worked answer

R=mg\cos30^\circ

F=mg\sin30^\circ

mg\sin30^\circ=\mu mg\cos30^\circ

\mu=\tan30^\circ

Answer

\mu=0.577\quad\text{(3 s.f.)}

Use equality only at the limit

Definition

What is meant by limiting equilibrium?

Model answer: Equilibrium in which friction has reached its limiting value and the body is on the point of moving.

0\le F\le \mu R,\qquad F_{\max}=\mu R

Symbols

$F$ = actual friction while at rest (N)
$F_{\max}$ = limiting friction (N)
$\mu$ = coefficient of friction (none)
$R$ = normal reaction (N)

At rest but not limiting: $F<\mu R$ .
On the point of moving: $F=\mu R$ .
While sliding in these Paper 4 models, use $F=\mu R$ opposite the motion.

Compare required friction with available friction

Swipe left or right to see the whole diagram →

For a body at rest, compare the friction required for equilibrium with the maximum friction available. The actual friction is the required value, not automatically μR.

Worked example

Can the block remain at rest?

A 5 kg block rests on a plane inclined at 20° to the horizontal. The coefficient of friction is 0.4 and no other force acts.

R=50\cos20^\circ=46.98\text{ N}

F_{\max}=0.4R=18.79\text{ N}

F_{\text{needed}}=50\sin20^\circ

F_{\text{needed}}=17.10\text{ N}

The required 17.10 N is below the available 18.79 N, so the block can remain at rest. Its actual friction is 17.10 N up the plane.

Answer

F=17.1\text{ N}

The friction acts up the plane, and equilibrium is possible.

Key idea

The test for rest is “friction needed ≤ friction available”. Do not replace the needed friction by

\mu R

before checking.

Recalculate R when the applied force is angled

An upward component of a pull reduces $R$ and therefore reduces limiting friction. A downward component increases both.

Worked example

A pull changes the normal reaction

A 10 kg block is on a horizontal floor with $\mu=0.25$ . It is about to move right under a force $P$ acting 30° above the horizontal.

R=100-P\sin30^\circ

P\cos30^\circ=0.25R

P\cos30^\circ=25-0.125P

Answer

P=25.2\text{ N}\quad\text{(3 s.f.)}

Your turnIndependent check [4 marks]

A block is in limiting equilibrium on a plane inclined at 30° to the horizontal, with no force other than weight and contact from the plane. Find the coefficient of friction.

Show worked answer

R=mg\cos30^\circ

F=mg\sin30^\circ

mg\sin30^\circ=\mu mg\cos30^\circ

\mu=\tan30^\circ

Answer

\mu=0.577\quad\text{(3 s.f.)}