Pulleys and car problems

Configuration: Two masses m1 and m2 connected by a rope over a pulley; m1 sits on a horizontal surface with kinetic friction μ m1 g; m2 hangs (the driver). Tension T is the same on both sides of the rope.
Key ideas:
- Tension is an internal force when the two masses are treated as a single system.
- Friction on m1 opposes the motion of the system along the horizontal direction.
- Gravity acts on m2 as the driving force; vertical motion for m1 is not present, so y-forces on m1 do not affect the acceleration.

Mass 1 (on surface):
- Along the track: $T - \,\mu m1 g = m1 a$
Mass 2 (hanging):
- Downward: $m2 g - T = m2 a$
Friction magnitude: $Ff = \mu m1 g$
Normal on m1: $N = m_1 g$ (vertical equilibrium on m1; no vertical acceleration)
Note: The friction term uses the kinetic friction coefficient μ when the system is moving; if the system is at rest, static friction μ_s would apply and the equations would be inequalities.

Add the two equations to eliminate T:
- $m2 g - \mu m1 g = (m1 + m2) a$
Solve for a:
- ${a = \frac{(m2 - \mu m1)\,g}{m1 + m2}}$
Interpretation:
- If $m2 > \mu m1$, the acceleration is positive (m2 moves downward, m1 moves right).
- If $m2 < \mu m1$, the system would slow and could reverse direction (static friction might prevent motion depending on μ_s and whether the system ever starts moving).

Use mass 2 equation: $m2 g - T = m2 a\;\Rightarrow\; T = m2 g - m2 a$
Substitute the expression for a:
- $T = m2 g - m2 \left(\frac{(m2 - \mu m1) g}{m1 + m2}\right)$
Simplify:
- Numerator simplifies to $m1 m2 g + \mu m1 m2 g$
- Final result: $\boxed{\;T = \frac{m1 m2 (1 + \mu)\,g}{m1 + m2}\;}$
Optional check using mass 1 equation:
- From $T - \mu m1 g = m1 a$, substitute a to obtain the same T (consistency check).

Total external horizontal/vertical forces on the system: only the driving force $m2 g$ and the friction force $\mu m1 g$ act externally along the motion direction.
Net external force: $F{ ext{ext}} = m2 g - \mu m_1 g$
Total mass: $M = m1 + m2$
Acceleration: $a = \frac{F{ ext{ext}}}{M} = \frac{(m2 - \mu m1) g}{m1 + m_2}$
This matches the previous result and reinforces the idea that tension is internal to the two-mass system.

The expressions assume kinetic friction on the lower mass; if the system is just starting to move, use static friction μ_s and inequalities.
The sign convention matters: choosing the direction of motion consistently yields correct a and T.
The algebra can be cumbersome; the clean final forms are useful for quick checks:
- $a = \dfrac{(m2 - \mu m1) g}{m1 + m2}$
- $T = \dfrac{m1 m2 (1 + \mu) g}{m1 + m2}$

Setup: A block of mass m on an incline of angle θ. Weight components:
- Along incline (downslope): $mg\sin\theta$
- Normal (perpendicular to incline): $N = mg\cos\theta$
Static friction force has maximum value: $F{f,\,\text{max}} = \mus N = \mu_s m g \cos\theta$
At the threshold of slipping (just about to move): balance along incline:
- $mg\sin\theta = \mu_s m g \cos\theta$
- Therefore: $\tan\theta = \mu_s$
- And the threshold angle: $\boxed{\theta = \arctan(\mu_s)}$
Numeric examples
- Example A: μ_s = 0.5 → $\theta = \arctan(0.5) \approx 26.6^{\circ}$
- Example B: Rubber vs concrete with μ_s = 1 → $\theta = \arctan(1) = 45^{\circ}$
Practical interpretation
- If the incline angle is below the threshold, the block does not slide (static friction can adjust up to μ_s N).
- If the angle exceeds the threshold, the block slides and kinetic friction μk acts (often ≈ μs for many materials, but not guaranteed).

Setup: A car of mass m travels on a flat circular path of radius r. Static friction provides the centripetal force.
Required centripetal acceleration: $a_c = \frac{v^2}{r}$
Maximum frictional (centripetal) force: $F_f = \mu N = \mu m g$ (N = mg on a horizontal surface)
Set centripetal acceleration equal to available friction per unit mass: $\frac{v^2}{r} = \frac{F_f}{m} = \mu g$
Solve for maximum speed: $\boxed{v_{\max} = \sqrt{\mu g r}}$
Numerical example
- Given: $r = 100\,\text{m}$, $\mu = 1$, $g \approx 9.8\,\text{m/s}^2$ →
- $v_{\max} = \sqrt{(1)(9.8)(100)} = \sqrt{980} \approx 31.3\,\text{m/s}$
- This is about (≈ 70) mph (order-of-magnitude note; ~60–70 mph range depending on g and rounding)
Design intuition
- To increase safety or speed through a bend, you can either increase the radius of curvature or reduce speed.
- If you’re approaching a corner too tightly, you’d want a larger-radius path (blue vs red analogy) to reduce required centripetal force and thus friction.
Important caveats
- If there is any banking or incline, normal force and friction directions change; here we assumed a flat road.

Friction coefficient μ is unitless; distinguish μs (static) vs μk (kinetic) in problems involving impending or ongoing motion.
Normal reaction on an incline: $N = mg\cos\theta$ .
Static friction limit on an incline: $Ff \le \mus N = \mus m g \cos\theta;$ at the threshold of motion, $mg\sin\theta = \mus m g \cos\theta$ and thus $\tan\theta = \mu_s$ .
For two-mass systems on a friction surface, treating the system as a single body gives a convenient route to the acceleration: $a = \frac{(m2 - \mu m1) g}{m1 + m2}$ (kinetic friction assumed).
Tension in the rope for the two-mass problem with kinetic friction is:
- $T = \frac{m1 m2 (1 + \mu) g}{m1 + m2}$
Centripetal motion with friction on a flat curve obeys: $\frac{v^2}{r} = \mu g \Rightarrow v_{\max} = \sqrt{\mu g r}$ and the equivalent radius form: $r = \frac{\mu g}{v^2}$ .
Real-world relevance includes parking on slopes, road safety around curves, and understanding everyday friction phenomena (e.g., refrigerator friction story mentioned as motivation).
Note on problem-solving approach:
- Start with clear free-body diagrams (especially for incline and two-mass pulley problems).
- Use consistent direction conventions; derive a and T step-by-step, then look for algebraic simplifications.
- When possible, present results in clean forms to avoid algebra traps (as highlighted when instructors suggest simpler expressions).