Deals with the study of forces and their effect on motion.
Newton's Second Law
F=ma is the fundamental equation.
F represents force.
m represents mass.
a represents acceleration.
Applies to a point mass or an object that can be treated as a point mass.
Non-inertial Reference Frames
Lab reference frame is considered a non-inertial reference frame.
In a non-inertial reference frame, the equation Fnet=ma does not hold true directly. A fictitious force needs to be introduced.
Fictitious Force: Ffictitious=−mA, where A is the acceleration of the non-inertial reference frame.
The transformation of acceleration from an inertial frame to a non-inertial frame is given by: a=a′+A, where:
a is the acceleration in the inertial frame.
a′ is the acceleration in the non-inertial frame.
A is the acceleration of the non-inertial frame with respect to the inertial frame.
Therefore, a′=a+(−A)=a−A
The net force in the inertial frame is Fnet=m(a′+A)
In the non-inertial frame, the modified Newton's second law is:
Fnet+(−mA)=ma′
F<em>net+F</em>fictitious=ma′
If one were to use the non-intertial reference frame, then Fnet=ma′
Example: Pendulum in an Accelerating Car
Consider a pendulum inside a car accelerating with acceleration A.
The tension in the string is T.
In the non-inertial frame (inside the car), the forces acting on the pendulum bob are:
Tension T.
Weight mg.
Fictitious force −mA.
The pendulum will deflect at an angle such that the net force balances.
Tx=mA
Ty=mg
The time period of the pendulum is T=2πg<em>effl, where g</em>eff is the effective acceleration due to gravity.
Here, geff=g2+A2
Steps to Solve F=ma Problems
Applicable to n-object systems.
Write down F=ma for each object. This will give you n equations.
Identify all unknown variables (n+m in total).
Identify all forces acting on each object.
Identify constraints on the motion of the objects. These constraints will give you m additional equations.
Solve the system of equations to find the unknowns.
Example: Two Blocks Connected by a String
Two blocks with masses M and m are connected by a string over a pulley.
M is on a horizontal surface, and m is hanging vertically.
Forces on M: Tension T and friction.
Forces on m: Tension T and weight mg.
Equations for M:
∑F<em>x=T=Ma</em>x
∑Fy=N−Mg=0
Equations for m:
∑F<em>y=mg−T=ma</em>y
Constraint: a<em>x=a</em>y=a
Solve for T and a.
Example: Block on an Inclined Plane in an Accelerating System
A block of mass m is on an inclined plane (angle θ) inside an accelerating system (acceleration A<em>x). The acceleration in y direction is zero (A</em>y=0).
The acceleration in the non-inertial frame is a=a′+A
a<em>x=a′</em>x+(−Ax)
a<em>y=a′</em>y+0
Forces: Normal force N, weight mg.
Equations:
Nsin(θ)=max (∑Fx = max)
Ncos(θ)=mg (∑Fy = may)
Constraint Example: Two masses connected by a string over a pulley.
m1 and m2 are connected by the string the constraints that relate the acceleration is a1=−a2