## Newton's 2nd law

Let us assume that our reference frame is an inertial frame.

Newton's second and third laws are valid in all inertial reference frames.

Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass.

Unbalanced forces cause acceleration.
no net force <--> no acceleration

The net force is the vector sum of all forces acting on the object.

F = ∑iFi,  Fx = ∑iFix,  Fy = ∑iFiy,  Fz = ∑iFiz

In algebraic form we write Newton's second law as F = ma

This is a vector equation.  It is equivalent to the three equations, Fx = max, Fy = may, Fz = maz.  The acceleration a = F/m is in the direction of the force and proportional to the magnitude of the force.  The mass of an object is a measure of its inertia, its resistance to changing its state of motion.  If two objects are supposed to have the same acceleration, then the more massive object must be acted on by a larger force, while the less massive object must be acted on by a smaller force.  The mass is a scalar quantity.

Units: In SI units, mass is measured in kg, acceleration in m/s2 and force in Newton (N).  1 N = 1 kg·m/s2.
In the British engineering system the unit of force is the pound (lb).  (Conversion: 1 pound-force = 4.448 N)

Given the same push or pull, larger masses accelerate less than smaller masses.  You need much less force to accelerate a tricycle than to accelerate a car.  Because of its inertia, you need a force to accelerate an object.  If there is no net force acting on an object, it will not accelerate, its velocity will not change.  If it is initially at rest, it will stay at rest, if it is moving with a given speed in a certain direction, it will keep on moving with the same speed in the same direction.

#### Problem:

A net force F, applied to an object of mass m1, produces an acceleration of 3 m/s2.  The same force, applied to a second object of mass m2 produces an acceleration of 1 m/s2.
(a)  What is the value of the ratio m1/m2?
(b)  If m1 and m2 are combined, find their acceleration under the action of the force F.

Solution:
(a) We have F = m1a1 and F = m2a2.  Therefore m1a1 = m2a2, m1/m2 = a2/a1 = 1/3.
(b) Now F = (m1 + m2)a = (4/3)m2a, since m1 = (1/3)m2
Therefore (4/3)m2a = m2a2, a = (3/4)a2 = 0.75 m/s2.

#### Problem:

A car with a mass of 850 kg is moving to the right with a constant speed of 1.44 m/s.
(a) What is the total force on the car?
(b) What is the total force on the car if it is moving to the left with a constant speed of 1.44 m/s.

Solution:
(a) The car moving to the right with constant speed is moving with constant velocity.  The acceleration is zero.  The total force on the car is zero.
(b) The car moving to the left with constant speed is moving with constant velocity.  The acceleration is zero.  The total force on the car is zero.

#### Problem:

A plot of velocity versus time for a m = 0.1 kg particle moving along the x-axis starting at the origin is shown below.
(a)  What is the net force acting on the particle at t = 6 s?
(b)  What is the net force acting on the particle at t = 12 s?
(c)  What is the net force acting on the particle at t = 16 s?

Solution:
Remember:  For one-dimensional problems the sign of a vector is the direction indicator.
(a)  Between t = 5 s and t = 10 s the velocity of the particle increases at a constant rate from 2 m/s to 7 m/s.
The acceleration of the particle is a = (5 m/s)/(5 s) = 1 m/s2 at any time between 5 s and 10 s.
The net force acting on the particle is F = ma = (0.1 kg)*(1 m/s2) = 0.1 N pointing in the positive x-direction.
(b)  At any time between 10 s and 15 s the velocity of the particle is constant, its acceleration is zero, the net force acting on the particle is zero.
(c)  Between t = 15 s and t = 19 s the velocity of the particle decreases at a constant rate from 7 m/s to 5 m/s.
The acceleration of the particle is a = (-2 m/s)/(4s) = -0.5 m/s2 at any time between 15 s and 19 s.
The net force acting on the particle is F = ma =( 0.1 kg)*(-0.5 m/s2) = -0.05 N.  F is pointing in the negative x-direction.

#### Problem:

A 3 kg mass undergoes an acceleration given by a = (2i + 5j) m/s2.  Find the resultant force F and its magnitude.

Solution:
F = ma, F = [6i + 15j] N is the resultant force.
F2 = Fx2 + Fy2 = (36 + 225) N2 = 261 N2,
F = 16.16 N is the magnitude of the resultant force.

#### Problem:

A heavy freight train has a mass of 15000 metric tons.  If the locomotive can exert a pulling force of 750000 N, how long does it take to increase the speed from 0 to 80 km/h?

Solution:
For an object starting from rest with constant acceleration a we have v = at, t = v/a. The magnitude of the acceleration of the train is
a = F/m = 750000 N/(15000*1000 kg) = 0.05 m/s2.  Therefore
t = (80000 m/3600 s)/(0.05 m/s2) = 444 s = 7.4 min.

#### Problem:

Suppose a truck loaded with sand accelerates at 0.5 m/s2 on a highway.  If the driving force on the truck remains constant, what happens to the truck's acceleration, if its trailer leaks sand at a constant rate through a hole in its bottom?

Solution:
F is constant.  F = ma.  As the mass decreases, the acceleration increases.
a = F/m,  da/dt = -F(1/m2) dm/dt = -(a/m) dm/dt.
If dm/dt is constant and negative, i.e. dm/dt = -b, with b a positive constant, then da/dt = (a/m)b.  The mass m is decreasing and a and da/dt are increasing with time.
We have m = m0 - bt,  da/dt = ab(m0 - bt).