How Hard Can Friction Push Back?
A block rests on a rough horizontal surface. A gentle push does not move it: friction simply matches the push. Push harder and friction grows too — until it reaches its ceiling $\mu R$. Beyond that ceiling the block slides. Set the weight and the push and watch friction follow, then hit the limit.
normal reaction $R$ = 40.0 N · limiting friction $\mu R$ = 16.0 N · friction $F$ = 6.0 N
Below the limit friction takes whatever value keeps the block in equilibrium, so $F \le \mu R$. At the limit $F = \mu R$.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. Friction between two surfaces acts:
R2. A block of weight $W$ rests on a rough slope at angle $\theta$. The normal reaction is:
R3. A block is on the point of moving but has not yet moved. It is:
Friction is the part of the contact force acting along the surface. It is self-adjusting: it takes whatever value is needed to prevent sliding, up to a maximum:
| Situation | Friction |
|---|---|
| At rest, not about to move | $F$ takes the value needed for equilibrium, with $F < \mu R$ |
| Limiting equilibrium — just about to move | $F = \mu R$ |
| Moving (sliding) | friction is at its maximum, $F = \mu R$ |
The common slip is to write $F = \mu R$ in every problem. It only applies at the limit.
$\mu$ is a number with no units. It depends on the pair of surfaces, not on the size of the block or the speed. Because $F = \mu R$, doubling the normal reaction doubles the friction available.
This gives the normal reaction $R$. Never assume it equals the weight.
If the body is on the point of moving, $F = \mu R$. Otherwise treat $F$ as an unknown.
On a rough slope where only the weight acts and the block is on the point of sliding, $W\sin\theta = \mu W\cos\theta$, so $\tan\theta = \mu$. This is why a steeper slope makes sliding more likely.
Groups of 4 use the inclined-plane lab. Phase 1: move both sliders and note R and the downhill component. Phase 2: fix W and increase theta — the downhill component grows while R falls. Phase 3: answer the auto-graded challenge on limiting friction and F = μR.
The weight, normal reaction and an applied force are already drawn. Use the sliders to supply the friction force that keeps the block in equilibrium. The dashed red resultant shrinks to zero when your friction is right, then answer the challenge on F = μR.
A block of mass $10\ \text{kg}$ rests on a rough horizontal surface. A horizontal force of $12\ \text{N}$ is just enough to move it. Find the coefficient of friction.
"Just enough to move it" means limiting equilibrium, so friction has reached $\mu R$. Resolve vertically first to find $R$.
The surface is horizontal with no other vertical force, so $R = W = 10\times 9.8 = 98\ \text{N}$.
At the point of moving, $F = \mu R$ with $F = 12\ \text{N}$, so $12 = 98\mu$.
$\mu = \dfrac{12}{98} = 0.122$ (3 s.f.).
A block of weight $40\ \text{N}$ rests on a rough slope inclined at $20^\circ$ to the horizontal. The coefficient of friction is $0.5$. Will the block slide? If it does not, find the friction force.
Compare the downhill component of the weight with the maximum friction available $\mu R$. Do not assume friction equals $\mu R$ unless sliding is about to happen.
$R = 40\cos 20^\circ = 37.6\ \text{N}$.
$\mu R = 0.5\times 37.6 = 18.8\ \text{N}$.
$W\sin 20^\circ = 40\times 0.3420 = 13.7\ \text{N}$. Since $13.7 < 18.8$, the block does not slide. Friction takes the value needed for equilibrium, $F = 13.7\ \text{N}$ acting up the slope — well below $\mu R$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. Which statement about friction is correct?
Q2. A block of weight $50\ \text{N}$ rests on a rough horizontal surface with $\mu = 0.3$. The limiting friction is:
Q3. A block of weight $60\ \text{N}$ is at rest on a rough horizontal surface with $\mu = 0.5$. A horizontal push of $8\ \text{N}$ is applied and the block does not move. The friction force is:
Q4. A block just begins to slide down a rough slope under its own weight. If $\mu = 0.6$, the angle of the slope is:
Work each on paper, then reveal the solution to check your working.
A block of mass $20\ \text{kg}$ rests on a rough horizontal floor with $\mu = 0.25$. Find the least horizontal force needed to move it.
Reveal Solution
$R = 20\times 9.8 = 196\ \text{N}$, so the least force is the limiting friction $\mu R = 0.25\times 196 = 49\ \text{N}$.
A block of weight $80\ \text{N}$ rests on a rough horizontal surface. A horizontal force of $20\ \text{N}$ is just enough to move it. Find $\mu$.
Reveal Solution
$R = 80\ \text{N}$ and at the limit $20 = \mu\times 80$, so $\mu = 0.25$.
A box of weight $100\ \text{N}$ is pulled at steady speed across a rough horizontal surface by a rope making $30^\circ$ with the horizontal. The coefficient of friction is $0.4$. Find the tension in the rope.
Reveal Solution
Resolve vertically: $R = 100 - T\sin 30^\circ$, so $F = 0.4(100 - 0.5T) = 40 - 0.2T$. Steady speed means $T\cos 30^\circ = F$, so $0.8660T = 40 - 0.2T$. Hence $1.0660T = 40$ and $T = 37.5\ \text{N}$.
A block of mass $8\ \text{kg}$ is at rest on a plane inclined at $20^\circ$ to the horizontal, held by a string parallel to the plane with tension $13\ \text{N}$. Find the frictional and normal components of the contact force.
Reveal Solution
$W = 8\times 9.8 = 78.4\ \text{N}$. Perpendicular to the plane $R = 78.4\cos 20^\circ = 73.7\ \text{N}$. Parallel to the plane the weight has component $78.4\sin 20^\circ = 26.8\ \text{N}$ downhill, so friction acts uphill with magnitude $F = 26.8 - 13 = 13.8\ \text{N}$.
A block of mass $5\ \text{kg}$ rests on a rough slope and is just on the point of sliding down it when the slope is inclined at $30^\circ$ to the horizontal. Find $\mu$, and the friction and normal reaction acting on the block.
Reveal Solution
At the limit $\tan\theta = \mu$, so $\mu = \tan 30^\circ = 0.577$. $W = 5\times 9.8 = 49\ \text{N}$ and $R = 49\cos 30^\circ = 42.4\ \text{N}$. Then $F = \mu R = 0.5774\times 42.4 = 24.5\ \text{N}$, which matches the downhill component $49\sin 30^\circ = 24.5\ \text{N}$.
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