Friction and Newton's Second Law: Forces and Applications

Friction: Nature and Types

Introduction to Friction

Friction is a force that arises when two surfaces are in contact and attempt to move relative to each other. It acts to prevent or resist sliding, playing a crucial role in everyday phenomena such as walking and driving.

• Static friction: Prevents the initiation of sliding between surfaces.

• Kinetic friction: Resists motion once sliding has begun.

• Friction depends on the normal force applied between surfaces.

• It is caused by the roughness of both surfaces and is measured by the coefficient of friction ($\mu$).

• Friction is independent of contact area and surfaces do not significantly deform under normal conditions.

Static Friction

Static friction acts to prevent motion when surfaces in contact attempt to move relative to each other. It balances the applied force up to a maximum value.

• Force of static friction ($\vec{f}_s$) acts with enough force to prevent motion.

• When the net force is zero, the object remains at rest:

$\vec{F}_{net} = \vec{f}_s + \vec{F}_{applied} = 0$

• Static friction balances the applied force:

$\vec{f}_s = -\vec{F}_{applied}$

• If the applied force exceeds the maximum static friction, sliding begins:

$f_{max,s} = \mu_s F_N$

• $\mu_s$ is the coefficient of static friction.

Kinetic Friction

Kinetic friction acts to resist motion when surfaces in contact are moving relative to each other. It is generally less than the maximum static friction.

• Force of kinetic friction ($f_k$) is constant during motion:

$f_k = \mu_k F_N$

• $\mu_k$ is the coefficient of kinetic friction.

• Typically, $\mu_s \geq \mu_k$ (harder to start moving than to keep moving).

• The direction of the applied force can change the normal and friction forces.

Applications of Friction

Friction is responsible for many practical phenomena, such as walking and the acceleration of vehicles.

• Walking: Friction between the ground and feet prevents slipping and enables forward motion.

• Driving: Friction between tires and the road allows cars to accelerate and change direction.

Example: When a person walks, static friction pushes the person forward as the feet adjust and no slipping occurs. For a car, as the tire speed increases and moves left, road friction resists motion and pulls the tire to the right, enabling acceleration.

Concept Check: Static and Kinetic Friction

Consider trials where a block is pushed with increasing force:

• Trial 1: Pushed with 10 N, block does not move.

• Trial 2: Pushed with 25 N, block does not move.

• Trial 3: Pushed with 26 N, block accelerates right.

Key Points:

• Static friction in Trials 1 and 2 is equal in magnitude to the applied force.

• Kinetic friction applies a constant force once sliding begins.

• Maximum static friction is between 25 N and 26 N.

Trial	Applied Force (N)	Block Motion	Friction Type
1	10	No motion	Static
2	25	No motion	Static
3	26	Accelerates	Kinetic

Newton's Second Law: Forces and Acceleration

Newton's Second Law in Vector Form

Newton's Second Law relates the net force acting on an object to its acceleration in each coordinate direction. It is a fundamental principle for analyzing motion.

• Vector equation:

$\vec{F}_{net} = m_s \vec{a}_s$

• Component form:

$F_{net,x} = m_s a_{s,x}$ $F_{net,y} = m_s a_{s,y}$

• Solving for acceleration:

$a_{s,x} = \frac{F_{net,x}}{m_s}$ $a_{s,y} = \frac{F_{net,y}}{m_s}$

• Net force is the vector sum of all forces acting on the object:

$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2$

Example 1: Calculating Acceleration

A 15 kg sled is pushed left with an 80 N force at 30° below the horizontal. The ground supports the sled with a 187 N normal force. The coefficient of kinetic friction is 0.1. Find the sled's acceleration.

• Step 1: Draw a labeled force diagram.

• Step 2: Identify forces:

Force	Expression	Value
Gravity	$F_G = -mg$	$-147N$
Normal	$F_N$	$+187N$
Kinetic Friction	$f_k = \mu_k F_N$	$18.7N$
Person (x-component)	$F_{P,x} = -80N \cos 30°$	$-69.28N$
Person (y-component)	$F_{P,y} = -80N \sin 30°$	$-40N$

• Step 3: Find net force components:

$F_{net,x} = F_{P,x} + f_{k,x} + F_{G,x} + F_{N,x} = -69.28N + 18.7N + 0 + 0 = -50.58N$

$F_{net,y} = F_{P,y} + f_{k,y} + F_{G,y} + F_{N,y} = -40N + 0 + (-147N) + 187N = 0$

• Step 4: Calculate acceleration:

$a_x = \frac{F_{net,x}}{m} = \frac{-50.58N}{15kg} = -3.37\,m/s^2$

$a_y = \frac{F_{net,y}}{m} = 0$

Result: The sled accelerates left at $-3.4\,m/s^2$.

Example 2: Static Equilibrium and Tension

An 80 kg chandelier hangs motionless from the ceiling using two ropes angled at 60° with the ceiling. Find the tension in each rope.

• Step 1: Draw a labeled force diagram.

• Step 2: Identify forces:

Force	Expression
Gravity	$F_G = -mg = -784N$
Rope 1 (x-component)	$F_{1,x} = +0.5F_1$
Rope 1 (y-component)	$F_{1,y} = +0.87F_1$
Rope 2 (x-component)	$F_{2,x} = -0.5F_2$
Rope 2 (y-component)	$F_{2,y} = +0.87F_2$

• Step 3: Apply equilibrium conditions:

Sum of horizontal forces: $F_{1,x} + F_{2,x} = 0$

Sum of vertical forces: $F_{1,y} + F_{2,y} - 784N = 0$

• Step 4: Solve for tension:

Since the angles are equal, $F_1 = F_2 = T$

$0.87T + 0.87T = 784N$

$T = \frac{784N}{2 \times 0.87} = 450.6N$

Result: Each rope supports a tension of approximately $451N$.

• Equal angles lead to equal tension magnitudes.

• Angled ropes require greater tension than vertical ropes to support the same weight.

• Vertical components support the weight; horizontal components cancel each other.

Concept Check: Static Equilibrium

• The tension in rope 1 and rope 2 is equal.

• The horizontal force components of rope 1 and 2 are equal in magnitude but opposite in direction.

• The upward vertical force components of rope 1 and 2 together are equal to the downward force of gravity.