1. 1. The following diagram shows a system of forces acting on a particle in a plane. A third force is added so that the particle rests in equilibrium. Find the magnitude of this force and the angle that make with the horizontal. 30  8 N 12 N Components diagram 8 cos 30 F cos  8 sin 30 F sin  12 Horizontal component:  8 cos 30 + F cos   12 = 0  F cos  = 12  8 cos 30 [1] Vertical component:  :8 sin 30  F sin  = 0  F sin  = 8 sin 30 [2]

2. Divide [2] by [1]:   = 38.3 Sub  into [2]: Added force is 6.45 N, acting at an angle of 38.3  below the horizontal.  F = 6.45 N F sin 38.3 = 4

3. 2. A particle is acted on by a force of 15 N which acts on a bearing of 020º, and another force of 4 N which acts on a bearing of 230º. Find the magnitude of a third force which will keep the system in equilibrium, stating the angle of its line of action as a bearing. θ 230º 20º North 4 N Let the third force be P N acting at θ to the horizontal, as shown. The force diagram is: P N

4. Components diagram Psin θ 4 cos 40 4 sin 40 Pcos θ Horizontally: (1) Vertically: (2) From (1): From (2): (3) (4) (4) divided by (3) Substitute into (3) P acts on a bearing of 190º and has size 15 cos 70 – P cos θ – 4 cos 40 = 0 15 sin 70 – P sin θ – 4 sin 40 = 0 P cos θ = 15 cos 70 – 4 cos 40 P sin θ = 15 sin 70 – 4 sin 40 P = 11.7 N

5. 3. The diagram shows a particle suspended from a horizontal beam by two unequal, light and inextensible strings. Given that the tension in the left string is 8 N and it makes an angle of 40º to the beam, and the other string makes an angle of 60º to the beam, find the tension in the other string and the mass of the particle.

6. Mg Components diagram T sin60 T cos60 Resolving horizontally: Resolving vertically: (1) (2) 8 sin 40 + T sin 60 – Mg = 0 Force diagram

7. Resolving horizontally: Resolving vertically: (1) (2) From (1): Substitute in (2): The mass of the particle is 1.61 kg. 8 sin 40 + T sin 60 – Mg = 0 The tension in the other string is 12.3 N

8. 4. A car of mass 1500 kg is broken down on a rough plane inclined at an angle of θ° to the horizontal, where sin -1  = ( 7 / 25 ). It is being pulled up the plane by means of a towrope, which is acting at 9º above the line of greatest slope of the plane. The resistance between the plane and the car has a magnitude of 1000 N. The car is at rest in equilibrium and is about to move up the plane. Find the tension in the towrope and the magnitude of the force of the plane on the car. 1500g T 1500kg Force diagram 9º9º

9. T cos9 1500g cos  1000 1500g sin  R n T sin9 Components diagram Res parallel: T cos9 – 1000 – 1500g  (7/25) = 0 T = 5180 N Res perpendicular: R n + 5100  sin9 – 1500g  (24/25) R n = 13300 N

10. 5. A light inextensible string has its ends attached to two fixed points A and B. The point A is vertically above B. A smooth ring R of mass m kg is threaded on the string and is pulled by a force of magnitude 3N acting upwards at 400 to the horizontal. The section AR of the string makes an angle of 300 with the downward vertical and the section BR is horizontal (see diagram). The ring is in equilibrium with the string taut. 2 N 30  45  mg kg R A B (i)Find the tension in the string. (ii)Find the value of m.

11. mg Components diagram 2 sin45 2 cos45 T Resolve  Horizontally: T + T cos 60 = 2 cos 45  T = 0.942809.. Resolve vertically: T sin 60 + 2 sin 45 = mg  m = 0.228 kg (a) T = 0.943 N(b) m = 0.228 kg