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Topic 6 of 20

Applications of Newton's Laws

Once Newton's Second Law is in hand, a systematic recipe — draw a free body diagram, choose axes, write $\sum F = ma$ in each direction — unlocks an enormous variety of mechanics problems. This topic applies that recipe to friction, ropes, pulleys, and uniform circular motion.

Key Concepts

Static Friction

Friction that prevents two surfaces from sliding. It adjusts to match the applied force up to a maximum:

f_s \leq \mu_s N

. When the applied force exceeds this maximum, the object begins to slide.

Kinetic Friction

Friction acting on already-sliding surfaces:

f_k = \mu_k N

, directed opposite to motion.

\mu_k < \mu_s

always — it takes more force to start sliding than to keep sliding.

Tension

Force transmitted through a rope or cable, directed away from the object along the rope. For a massless, inextensible string, tension is the same at every point.

Centripetal Acceleration

For uniform circular motion (constant speed around a circle), acceleration points toward the center:

a_c = v^2/r = \omega^2 r

. The net centripetal force is

F_c = mv^2/r

Atwood Machine

Two masses

m_1

and

m_2

connected by a string over an ideal pulley. The heavier mass accelerates downward at

a = (m_1-m_2)g/(m_1+m_2)

. Classic example of treating a two-body system.

Inclined Plane

On a frictionless incline at angle

\theta

: the component of gravity along the slope is

mg\sin\theta

, giving

a = g\sin\theta

. With friction:

a = g(\sin\theta - \mu_k\cos\theta)

sliding down.

Key Equations

Static friction (maximum)

f_{s,\max} = \mu_s N

Friction force can be anything from 0 to this maximum; object stays put as long as F < f_s,max.

Kinetic friction

f_k = \mu_k N

Always this exact value when surfaces are sliding. Directed opposite to motion.

Centripetal force

F_c = \frac{mv^2}{r} = m\omega^2 r

Net force directed toward the center required for circular motion. Not a new type of force — it is provided by tension, gravity, normal force, friction, etc.

Atwood acceleration

a = \frac{(m_1 - m_2)\,g}{m_1 + m_2}

Acceleration of the system when m1 ≠ m2, ignoring pulley mass and friction.

Inclined plane acceleration

a = g\sin\theta - \mu_k g\cos\theta

Net acceleration down a rough incline; first term is the driving force, second is friction.

Worked Example

Block on a Rough Incline

Problem

A 4 kg block slides down a 30° incline with $\mu_k = 0.20$ . Find the acceleration (take $g = 10$ m/s²).

Solution

Draw FBD. Forces along the incline: gravity component down the slope, kinetic friction up the slope.

Along-slope equation ( $+$ down the incline):

\sum F = mg\sin\theta - f_k = ma

Normal force from the perpendicular equation: $N = mg\cos\theta$ . So $f_k = \mu_k mg\cos\theta$ .

a = g\sin 30° - \mu_k g\cos 30° = 10(0.5) - 0.20\times10\times0.866

a = 5.0 - 1.73 \approx 3.3 \text{ m/s}^2

Answer The block accelerates at ≈ 3.3 m/s² down the incline.

Practice

Exercises

7 problems

1 of 7

A $10 \text{ kg}$ block rests on a surface with $\mu_s = 0.4$ ( $g = 10 \text{ m/s}^2$ ). What is the maximum static friction force?

Your answer

2 of 7

The same block is now sliding with $\mu_k = 0.25$ ( $g = 10 \text{ m/s}^2$ ). What is the kinetic friction force?

Your answer

3 of 7

A $2 \text{ kg}$ object moves in a circle of radius $0.5 \text{ m}$ at a speed of $3 \text{ m/s}$ . What centripetal force is required?

Your answer

4 of 7

An Atwood machine has masses $m_1 = 3 \text{ kg}$ and $m_2 = 5 \text{ kg}$ ( $g = 10 \text{ m/s}^2$ ). What is the magnitude of the acceleration?

Your answer

m/s²

5 of 7

A $5 \text{ kg}$ block slides down a frictionless incline at $30°$ ( $g = 10 \text{ m/s}^2$ ). What is its acceleration?

Your answer

m/s²

6 of 7

A car rounds a flat curve of radius $50 \text{ m}$ at $20 \text{ m/s}$ . What minimum coefficient of static friction is needed? ( $g = 10 \text{ m/s}^2$ )

Your answer

(dimensionless)

7 of 7

A $60 \text{ kg}$ person is in an elevator decelerating downward at $2 \text{ m/s}^2$ ( $g = 10 \text{ m/s}^2$ ). What does a scale read?

Your answer

Key Takeaways

Friction is a contact force: static friction adjusts to prevent motion; kinetic friction is constant at $\mu_k N$ .
Always find the normal force before computing friction — $N \neq mg$ on inclines or when vertical forces are present.
Centripetal force is the net force directed toward the center; it is provided by existing forces, not a separate one.
For multi-body problems, either treat the system as a whole (to find $a$ ) or isolate each body (to find internal forces like tension).
On an incline, rotate the coordinate system so one axis is along the slope to simplify the equations.