Revision Note 5: Friction

Frictional force (more correctly, the maximum frictional force) is normal force times coefficient of static or dynamic friction (as the case may be).

F_fr = μN

The normal force (N) that is used to compute frictional force is best computed by considering the force by the surface on the body – in other words you need the free-body diagram of the body. This is the best method to compute the force by the body on the surface, as by Newton’s third law the two forces are numerically the same.

Problem Solving Tips:

Tip 5.1:

What is F_fr  for a ramp of incline θ to horizontal when the coefficient of friction is μ? By a previous tip (Tip 2.3), N = mg.cosθ. So,

F_fr,θ = μN = μ mg.cosθ

Tip 5.2: Up-down intuition:

Relating θ with normal force, N, and frictional force, F_fr,θ:

θ ↑        N ↓      F_fr,θ  ↓

The effect of friction decreases as the ramp becomes more and more vertical.

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 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.