1. TOPIC 9 The Equilibrium of Particles and Rigid Bodies

2. Resultant of Horizontal Forces A PARTICLE is a very small object, it has a mass but no size. A RIGID BODY has both a mass and a size. EQUILIBRIUM is when a particle or rigid body does not move when acted on by two or more forces. The RESULTANT of a number of forces is a single force which would produce exactly the same effect acting alone as the other forces would when acting together. Eg1 Eg 2 4N 7N 3N is equivalent to 2N10N 8N Is equivalent to

3. Resultant of Forces at an Angle The resultant of two forces at right angles to one another is represented by the diagonal of the rectangle whose sides represent the two forces. eg F = √7 2 + 2 2 Tan θ = 2 = √49 + 47 = √53θ = 15.95° = 7.28N θ 2N F 7N is equivalent to

4. Resultant of Forces at an Angle Since any two forces at right angles may be combined to produce a resultant, then any force can be resolved into two mutually perpendicular forces called the components of the force (we have already done this in Topic 5) A single force F at an angle θ to the horizontal can be split into the two mutually perpendicular forces Fcosθ in the horizontal direction and Fsinθ in the vertical direction. 7N is equivalent to 7sin27° = 3.18N 7cos27° = 6.24N 27°

5. Resultant of Forces at an Angle – Example 1 Demonstrate that the system of forces on the left is equivalent to the single force on the right. 16N 10N 8N 12N 60° 30° 40.4° 4.74N

6. Resultant of Forces at an Angle – Answer 1 Draw a table to find the resultant of the forces. ForceComponent in x-directionComponent in y-direction 10N O 16N0N-16N 8N8 cos 60° = 4N8 sin 60° = 6.928N 12N-12 cos 30° = -10.392N12 sin 30° = 6N Total3.608N-3.072N

7. Resultant of Forces at an Angle – Answer 1 The resultant is: F= √3.608 2 + 3.072 2 Tan θ = 3.072 = √13.02 + 9.44 3.608 = √22.46= 0.85 = 4.74N θ = 40.4° 3.608N -3.072N θ F

8. Resultant of Forces at an Angle – Example 2 A particle of mass 4kg is attached to the lower end of a light inextensible string. The upper end is fixed to a wall. A horizontal force of P newtons is applied to the free end of the string so that the string makes an angle of θ with the downward vertical and experiences a tension of 200N. If the 4kg mass rests in equilibrium, find P and θ.

9. Resultant of Forces at an Angle – Answer 2 Resolving vertically 200Cosθ = 4g Cosθ = 4gResolving horizontally 200200sinθ = P = 4 x 10200 x sin 78.46° = P 200200 x 0.978 = P = 0.2P = 195.96 N θ = cos -1 (0.2) θ = 78.46° 200 N 4g N θ P θ

10. Resultant of Forces at an Angle – Example 3 A particle of mass mkg is attached to the lower end of a light inextensible string. The upper end of which is fixed to a wall. A horizontal force of 40N is applied to the free end of the string so that the string makes an angle of θ with the downward vertical and experiences a tension of 90N. If the particle rests in equilibrium, find θ and m.

11. Resultant of Forces at an Angle – Answer 3 Resolving horizontally 90sinθ = 4090cosθ = mg sinθ = 4090 x cos26.39° = m x 10 9090 x 0.896 = 10m = 0.44480.64 = 10m θ = sin -1 (0.444)m = 80.64 θ = 26.39° 10 m= 8.06 kg 90 N mg θ 40N θ

12. Resultant of Forces at an Angle – Example 4 A body of mass 5kg is supported by two inextensible strings, the other ends of which are attached to two fixed points P and Q in a ceiling. The 5kg mass rests in equilibrium with one string experiencing a tension T newtons and inclined at 30° to the horizontal and the other experiencing a force of S newtons and inclined at 45° to the horizontal. Find T and S.

13. Resultant of Forces at an Angle – Answer 4 Resolving horizontallySubstitute 1 into 2 Scos45° = Tcos30°1.225T x cos45° +Tcos60° = 5g S = Tcos30°1.225T x 0.707 + T x 0.5 = 5x10 cos45°0.866T + 0.5T = 50 S = 1.225T1.366T = 50 T = 50 = 36.6N Resolving vertically 1.366 Scos45° + Tcos60° = 5g Use 1 to find S S = 1.225 x 36.6 = 44.8N S 5g 45°60° 30° 45° T