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Newton's Laws Of Motion & Friction

14. Angle of Friction

14. Angle of Friction (φ)
14.1Mathematical significance : The angle of friction (f) may be defined as the angle between the normal reaction N and the resultant of friction force f and the normal reaction.

Thus tan φ = f/N
since f = μN . therefore tan φ = μ
14.2 Physical Significance : The angle of repose (φ) is that minimum angle of inclination of the inclined plane at which a body placed at rest on the inclined plane is about to slide down in equilibrium condition.

N = mg cos θ and mg sin θ = f ≤ fmax
if θ = φ then f = fmax
∴ mg sin φ = fmax= μN
⇒ mg sin φ = μ mg cos φ
⇒ tan φ = μ
(a) When θ < φ (or tan–1μ) the body is in equilibrium
(b) When the angle of inclination is more than the angle of friction (θ > φ) the block starts sliding down.
14.3 Conditions for equilibrium of block (Depending upon the direction of applied force)
(a) Force parallel to the incline
(b) Force normal to the incline
(c) External Horizontal force
(a) Force parallel to the incline

(i) When θ > φ the friction force acting at its maximum value (fmax= μN) is incapable of keeping the block stationary. Therefore, parallel force is required to keep the block in equilibrium
(ii) The minimum value of this force is
(Fp)min = mg sin θ – fmax
(Fp)min = mg (sin θ – μ cos θ)
(iii) If a force slightly greater than (Fp)min is applied, then the block does not start moving up, but the force of friction force gets reduced.
(iv) It becomes to zero when the external force attains a value equal to Fp = mg sin θ. When the force is further increased (Fp> mg sin θ). The block has a tendency to move upward and direction of friction force gets reversed
(v) The block does not start moving up unless the external force attains the maximum value. The maximum value of Fpis given by

(Fp)max = mg sin θ + fmax = mg (sin θ + μ cos θ)
Conclusion :
The block remains stationary if (Fp)min ≤ Fp ≤ (Fp)max
mg(sin θ – μ cos θ) ≤ Fp ≥ mg (sin θ + μ cos θ)
(b) Force Normal to the incline (FN)

(i) Force FN applied normal to the inclined plane increases the magnitude of the frictional force by increasing the normal reaction.
(ii) Therefore N = mg cos θ + FN. When FN has its minimum value (FN) minthe friction force acting at its maximum value is just capable of preventing the block from sliding down.
(iii) That is
mg sin θ = fmax= μN
= μ [mg cos θ + (FN)min]
or (FN)min= m/μ g(sin θ – μ cos θ)
(iv) It is important to note that whatever may be the magnitude of FN, the block never attains a tendency to slide upward.
(v) When the magnitude of FN is more than its minimum value (FN)min then only the magnitude of friction forces decreases.
Conclusion :
The block remains stationary if

FN ≥ (FN)min ⇒ FN ≥ mg/μ (sin θ – μ cos θ)
(c) External Horizontal Force
It serves two purposes :
(i) It supports the frictional force
(ii) It increases the normal reaction and thus increases the magnitude of the limiting force of friction
Minimum Horizontal force (FH)min:-
(i) When the horizontal force acts as its minimum value, the friction force acts at its maximum value.
(ii) Applying the conditions of equation parallel and normal to the plane we get.

(FH)mincos θ = mg sin θ– fmax and (FH)minsin θ + mg cos θ = N and Fmax = μN
Solving these

Maximum Horizontal Force : (FH)max
(iii) As the magnitude of FH is slightly increased from its minimum value, the block does not start moving up, it remains stationary
(iv) But the magnitude of the friction force starts decreasing and it becomes equal to zero when FH= mg tan θ.
(v) If FH is further increased, the block has tendency to move upward and it just starts moving up when FH attains its maximum value.
(vi) From the free body diagram of the block
(FH)max cos θ = mg sin θ + fmax and (FH)max sin θ + mg cos θ = N
since fmax = μN therefore

Conclusion
The box remains stationary if
(FH) min ≤ FH ≤ (FH)max
≤ FH


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