Forces, Newton's Laws, Friction, and Circular Motion: Quizzes 4-6 Key Study Guide

Forces on Inclined Planes

Free-Body Diagrams and Force Components

When analyzing objects on inclined planes, it is essential to break down the forces acting on the object into components parallel and perpendicular to the surface.

• Normal Force (n): The force perpendicular to the surface, supporting the object against gravity.

• Frictional Force (fk): The force parallel to the surface, opposing motion.

• Weight (w): The gravitational force acting vertically downward.

Example: For a skier on a mountain inclined at 30° above the horizontal, the free-body diagram includes the normal force, friction, and weight. The normal force acts perpendicular to the slope, friction acts parallel and opposite to motion, and weight acts vertically downward.

Relative Magnitude of Forces

Comparing the magnitudes of forces acting on an object on an incline:

• Normal force (n) is typically less than the weight (w) due to the angle of the incline: $n = w_y$.

• Frictional force (fk) is proportional to the normal force: $f_k = bc_k n$.

• If $w > n$ and $bc_k < 1$, then $w > f_k$.

Additional info: The skier's motion and the coefficient of friction determine which force is greatest.

Newton's Laws of Motion

Newton's Third Law

Newton's Third Law states that for every action, there is an equal and opposite reaction.

• Action-Reaction Pair: If you push your sister on a skateboard, the force you exert on her is equal in magnitude and opposite in direction to the force she exerts on you: $F_{sonY} = F_{YonS}$.

• Net Force: The net force on each person depends on their mass and acceleration: $F_{net} = ma$.

Example: If your sister accelerates, her net force is not zero, while yours may be if you are stationary.

Jumping and Free Fall Motion

Calculating Initial Velocity and Force During Jump

When an object (e.g., a frog) jumps upward, the initial velocity required to reach a certain height can be calculated using kinematic equations.

• Initial velocity for vertical jump: $v_i = sqrt{2a_y Delta y}$

• Example calculation: For a jump to 1.0 m: $v_i = sqrt{2 times 9.8 times 1.0} = 4.427 text{m/s}$

• Force during jump: $a = frac{Delta v}{Delta t}$, so $a = frac{4.427}{0.15} = 29.51 text{m/s}^2$

Additional info: The force exerted by the frog on the ground can be found using $F = ma$.

Tension and Weight in Vertical Motion

Constant Velocity and Tension

When lifting an object (e.g., a piano) vertically at constant velocity, the tension in the rope equals the weight of the object.

• Constant velocity: $a = 0$, so $F_{net} = 0$

• Tension equals weight: $T = w = mg$

Additional info: If the object accelerates upward, tension must be greater than weight; if downward, less than weight.

Friction and Static Equilibrium

Static and Kinetic Friction

Frictional forces resist the relative motion of surfaces in contact.

• Static friction: $f_{s, max} = mu_s n$

• Kinetic friction: $f_k = mu_k n$

• Example: For two boxes, the maximum static friction is $f_{s, max} = mu_s 2mg$

Forces Between Vehicles and Rolling Friction

Action-Reaction Pairs and Rolling Friction

When two vehicles interact (e.g., a car and a truck), the forces they exert on each other are equal and opposite.

• Rolling friction: $f_r = mu_r N$

• Action-reaction: $F_{PonC} = F_{TonC}$

• Net force: If $a = 0$, $F_{net} = 0$ for both vehicles.

Additional info: The magnitude of forces can be compared based on mass and acceleration.

Circular Motion and Centripetal Acceleration

Centripetal Acceleration

Objects moving in a circle experience centripetal acceleration directed toward the center of the circle.

• Centripetal acceleration: $a_c = frac{v^2}{r}$

• Example: For $v = 10 text{m/s}$ and $r = 100 text{m}$, $a_c = frac{100}{100} = 1 text{m/s}^2$

Banked Curves and Maximum Speed

On a banked curve, the normal force and friction provide the necessary centripetal force to keep a car from slipping.

• Maximum speed without slipping: $v_{max} = sqrt{r g ( an heta + mu_s)}$

• Apparent weight: At the top of a curve, apparent weight may be less than actual weight due to centripetal acceleration.

Additional info: Apparent weight is the normal force felt by the person, which can be calculated as $N = mg - m a_c$ at the top of the curve.

Orbital Motion and Satellite Period

Geostationary Satellite Period

The period of a geostationary satellite is the time it takes to complete one revolution around the Earth, matching the Earth's rotation.

• Period calculation: $T = 24 H times 3600 text{s}/H = 86,400 text{s}$

• Application: This period ensures the satellite remains above the same point on Earth's surface.

Summary Table: Key Forces and Equations

Situation	Key Force	Equation
Inclined Plane	Normal Force	$n = mg cos heta$
Friction	Static/Kinetic Friction	$f_s = mu_s n$, $f_k = mu_k n$
Vertical Lift	Tension	$T = mg$
Circular Motion	Centripetal Acceleration	$a_c = frac{v^2}{r}$
Satellite Orbit	Period	$T = 86,400 text{s}$