Starting to push a heavy cupboard is harder than keeping it moving.
You push a heavy cupboard resting on the floor.
Takeaway: Static friction is usually larger than kinetic friction—hence the tough first shove.
The sofa resists motion chiefly because of _____ between its legs and the floor.
It acts before motion begins.
Yes—static friction holds objects at rest.
Not quite. Think before sliding starts.
Microscopic view: rough peaks, interlocking, and adhesion.
Even polished surfaces have tiny bumps called asperities.
Asperities fit together and resist relative motion until enough force breaks the locks.
Atoms at the contact points share electrons, creating adhesion that adds to friction.
Heavier loads press asperities closer, strengthening both interlocking and adhesion.
Use relative motion to tell them apart.
Usually \( \mu_s > \mu_k \). Decide relative motion first, then choose the correct formula.
Maximum static friction equals the coefficient of static friction multiplied by the normal reaction \(N\), which acts perpendicular to the surface.
Here, \( \mu_s \) is unit-less and typically lies between 0 and 1.
Source: NCERT Eq. 4.13
Kinetic friction opposes sliding motion. Its magnitude equals the product of the coefficient of kinetic friction and the normal reaction.
\(\mu_k\) stays nearly constant for a given surface pair at moderate speeds—use the relation to find friction on any steadily moving object.
Source: NCERT Eq. 4.14
Resolve forces and spot the friction direction.
The downhill pull \(mg\sin\theta\) must beat the maximum static friction to start motion.
Tip: Motion begins when \(mg\sin\theta = \mu_s N\). Increase θ to reach the “angle of repose.”
2 kg block on \(30^{\circ}\) incline, \(\mu_k = 0.2\). Find its acceleration.
Show \(mg\sin\theta\) down-slope, \(N\) normal, \(f_k\) up-slope.
\(N = mg\cos\theta = 2 \times 9.8 \times \cos30^{\circ} \approx 17 \text{ N}\).
\(f_k = \mu_k N = 0.2 \times 17 \approx 3.4 \text{ N}\).
\(F_{\text{net}} = mg\sin\theta - f_k \approx 9.8 - 3.4 = 6.4 \text{ N}\).
\(a = \frac{F_{\text{net}}}{m} = \frac{6.4}{2} = 3.2 \text{ m/s}^2\).
Use \(\mu_k\) because the block is moving; \(\mu_s\) applies only before motion starts.
A 10 kg crate rests on a level floor where the coefficient of static friction is 0.5. What horizontal force just starts it moving? Choose the closest value of \(f_s^{\text{(max)}}\).
Normal force \(N = mg\) on level ground.
Well done.
Re-calculate \(\mu_s N\).
Friction opposes relative motion of the two surfaces.
It is not always opposite to the object’s velocity.
Know these four points to recall the concept.
Starting friction is higher than sliding for most dry surfaces.
\(f_s^{\text{max}}=\mu_s N\), \(f_k=\mu_k N\).
Acts opposite to relative motion or its tendency.
Coefficients \(\mu\) vary with the surface pair.
On a level road, the car experiences forward friction during acceleration because:
Think action–reaction between tyre and road.
Exactly! The backward push of the tyre produces an equal forward friction force.
Remember Newton’s third law at the tyre–road contact.