Example (1): A horizontal force smaller than the maximum friction

• A box of weight 50 N rests on a horizontal floor whose coefficient of static friction with the box is 0.4.
• The box is pulled by a horizontal force P.
• Find the friction when P = 15 N, then find the greatest value of P for which the box remains at rest.
• Solution:

1. R = W = 50 N.
2. Fmax = μ R = 0.4 × 50 = 20 N.
3. from ΣX = 0 we get F = 15 N, which is less than 20 N, so the box remains at rest.
4. The greatest value of P is 20 N; at this value the box is on the point of moving.

Example (2): A pull inclined above the horizontal

• The same box (weight 50 N, μ = 0.4) is pulled by a force P that makes an angle of 30° above the horizontal.
• Find the greatest value of P for which the box remains at rest, giving the answer to two decimal places.
• Solution:

1. R = 50 − P sin 30° = 50 − P ÷ 2.
2. P cos 30° = μ R, that is P × √3 ÷ 2 = 0.4 (50 − P ÷ 2).
3. 0.866 P = 20 − 0.2 P, so 1.066 P = 20, giving P ≈ 18.76 N.
4. Notice that the inclined force gives a smaller limiting value of P (18.76 N) than the horizontal force (20 N), because part of the pull reduces the normal reaction.

Example (3): Finding the coefficient of friction

• A body of weight 40 N rests on a rough horizontal plane.
• When a horizontal force of 12 N is applied, the body is on the point of moving.
• Find the coefficient of static friction and the angle of friction.
• Solution:

1. R = W = 40 N.
2. F = P = 12 N.
3. μ = F ÷ R = 12 ÷ 40 = 0.3.
4. tan λ = 0.3, so λ ≈ 16.7°.

Example (4): A force inclined at 60° above the horizontal

• A body of weight 60 N rests on a rough horizontal plane with coefficient of static friction 0.5.
• It is pulled by a force P making an angle of 60° above the horizontal, and it is on the point of moving.
• Find P.
• Solution:

1. R = 60 − P sin 60° = 60 − P × √3 ÷ 2 ≈ 60 − 0.866 P.
2. P cos 60° = μ R, that is 0.5 P = 0.5 (60 − 0.866 P).
3. 0.5 P = 30 − 0.433 P, so 0.933 P = 30, giving P ≈ 32.15 N.
4. R ≈ 60 − 0.866 × 32.15 ≈ 32.15 N, the friction is 0.5 × 32.15 ≈ 16.08 N, and the horizontal component is 32.15 × 0.5 ≈ 16.08 N; the two are equal.