Revision Note 4: Newton’s Second Law

Newton’s Second Law:

F_net = ma

Here,

·        F_net = the net force (sum of all forces) on the body

·        m = the mass of the body

·        a = the acceleration of the body

When F_net = 0, a = 0 Newton’s second law becomes equivalent to his first law.

Problem Solving Tips:

Tip 4.1:

The following diagram relates the techniques in Revision Notes 2 (Static equilibrium), Revision Note 3 (Kinematics) and Revision Note 4 (Newton’s second law):

Tip 4.2:

When multiple bodies are pulled in a chain (see picture), then Newton’s second law is satisfied by (i) both bodies together, (ii) Body A, (iii) Body B.

In the example below, F = 2N, T = 1N. Where did the extra 1N force go? It goes to feed the acceleration of Body B. The ‘rough’ statement of this:  Acceleration eats Force.

Tip 4.3:

Impulse is defined as F.∆t, and that is equal to change in momentum:

F.∆t = ∆p

Tip 4.4:

These vectors have the same directions: F, ∆p, ∆v, and that direction is unrelated to the directions of v, p. Note that ∆v and v may not have the same direction. In Physics, v is not so important, ∆v is.

Revision Note 3: Kinematics

The following three equations of kinematics are valid only when acceleration is constant:

1.      v(t) = v(0) + at

2.      x(t) = x(0) + v(0)t + ½ at²

3.      [v(t)]² = [v(0)]² + 2a[x(t) – x(0)]

Here:

·        Time = t

·        Acceleration = a

·        Velocity at time t = v(t)

·        Position at time t = x(t)

Motion in 2D
The x- and y-directions have separate kinematics equations that do not interact. Only time connects the two.  So, the position of the particle at time t is (x(t), y(t)). If the particle is stopped (e.g., by falling to the ground), then the time of flight is the time when it hits the ground – generally this comes from the vertical kinematics equations.

If one direction (generally horizontal = x-direction) doesn’t have any acceleration, then the distance travelled in that direction is easy to find (x = vx . t). The total distance travelled in that direction then is velocity times time of flight.

Problem Solving Tips:

Tip 3.1:

When the acceleration is not constant it is expressed as a(t). Then use Calculus to find v(t), x(t) etc by integrating a(t).

Tip 3.2:

The time for a falling body to drop a height, h, is given by h = ½ gt².

Tip 3.3:

If you throw a particle vertically upwards with velocity, v, then the maximum height it reaches is given by the equation: v² = 2gh. This equation can be derived via kinematics and also from energy principles:

PE_i    +      KE_i       =    PE_f     +   KE_f

0       +   ½ mv²   =    mgh   +    0

        v²   =  2gh

Tip 3.4: What goes up …

What goes up in time, t, also comes down in time, t. In other words the equation from Tip 3.2 works in both directions – going up and going down. Not just that, the velocities going up and going down at any point in the flight have the same magnitude (the directions are opposites of each other).

Tip 3.5: Inclined planes

A particle pushed up an inclined plane (without friction) gains the same vertical height as a particle thrown vertically up. This follows from the energy principles as in Tip 3.3.

Even though the vertical heights are the same, the particle on the inclined plane travels a s longer distance and takes more time.

Tip 3.6: Up-down intuition:

Relating the height, h, from which a ball is dropped with the time, t, of flight:

h ↑ t ↑

The time goes up is less since for the latter part of the journey the speed of the ball is very high.

Relating the velocity, v, of a ball thrown up and the height, h:

v ↑ h ↑

Here too higher initial velocity cannot boost the height too much since the latter part of the journey is covered at lower speeds.

Tip 3.7: Projectile motion:

If v = initial velocity and θ = angle of launch (projectile launched from the ground level), then:

Maximum height = h = v².sin²θ / 2g

Range = R = v².sin 2θ / g

Time of flight = T = 2v.sin θ /g

The maximum value for h, T are achieved when θ=90 and for R when θ=45.

For maximum range: v_x = v_y at launch.

Revision Note 2: Static equilibrium

Conditions for static equilibrium:

1.       F_net = 0 (net force is zero)

2.       τ_net = 0 (net torque is zero)

For calculations:

·         Break F_net down into F_x, F_y (with proper signs) and then test for zeros along each axis.

·         For rotational moments, balance the clockwise and anti-clockwise moments.

Problem Solving Tips:

Tip 2.1:

In a force diagram, you will have to decide which direction is positive and which is negative. You can choose anything, but it helps to decide beforehand so that you do not waste time during the exam debating this.

Convention:

Vertically up is positive: E.g., g = -9.8 ms^-2.

Horizontally right is positive.

Along the Slope: x-axis, positive to the right.

Perpendicular to the slope: y-axis, positive upwards.

Tip 2.2:

Unknowns and signs: Suppose there is an unknown tension, T.

First, if you know its expected direction, draw it that way. Then use T or -T depending on the direction you chose. If you guessed the direction right, the numerical answer will be positive. If the numerical answer is negative, then the actual direction is in the reverse. It is like this: acceleration a = -5ms^-2 rightwards is actually +5ms^-2 leftwards.

Tip 2.3:

If θ is the angle of the inclined plane and a particle of mass m is on it, then the component of the weight along the x-axis and the y-axes are mg.sin θ and mg.cos θ respectively. Use the principle of extreme values to check this answer – when θ = 0 (plane is flat), then the component along the plane is 0 as it should be. This may seem counter to the “shadow = cos θ” tip, but is not since the θ is different. The correct angle (90 - θ) would match up with the shadow tip.

Tip 2.4: Up-down intuition:

Relating θ of inclined plane to a_x = component of g along the plane

θ ↑ a_x ↑

Intuition: When θ=90 then a_x = g, the highest possible value.