CHAPTER 2 – FORCE 2.1 Definition of a force, resolution of forces and resultant force A force may be defined as that which causes, or tends to cause, a change in the state of rest or the uniform motion, of a body. Two or more forces acting on a body can be replaced by a single force that is called a resultant force. Conversely, it is also possible to split or resolve one force into two forces. These forces are called the components of the original force. The process of splitting or resolving a force into its components is called resolution. 2.2 The Effects Of A Force A force can cause a stationary body to move or alter the speed and direction of a moving body. A change in velocity of a body is an acceleration. Therefore, a force causes an acceleration. Because both the magnitude and direction of a force are significant, forces are vector quantities. Types of forces in mechanics include weight, pushes and pulls, friction, normal reaction, tension and thrust. A force is characterised by: i) ii) iii) iv) its magnitude its direction its point of action the type of force 2.3 Newton’s Laws Newton’s first law of motion (Hukum Newton I): Newton’s first law of motion states that a body will continue in its state of rest or uniform motion unless a resultant external force acts on it. As a consequence: If a body is accelerating, there must be a resultant external force acting on it. If a body is not accelerating the forces acting upon it must be in equilibrium – the resultant external force must be zero. 1 Newton’s second law of motion (Hukum Newton II): Newton’s second law of motion states that the rate of change of momentum of a moving body is proportional to the resultant external force acting on the body and takes place in the direction of the force. When an external force acts on a body of constant mass, the acceleration produced is directly proportional to the force. ∑ F = ma Newton’s third law of motion (Hukum Newton III): Newton’s third law of motion states that if a body A exerts a force on a body B, then B exerts an equal and opposite force on A. This is sometimes stated as “ Action and reaction are equal and opposite “ FA A B FB FA (action) = FB (reaction) 2.4 Measurement of forces in newton The SI unit of force is the newton(N). One newton is the force required to give a mass of 1 kilogram an acceleration of 1 metre per second per second ( 1 m/s2 ). 1 N = 1 kg.m/s2 2 F = m a, therefore, 1N 1kg 1 m/s2 2N 1kg 2 m/s2 3N 1kg 3 m/s2 2.5 Conditions for Equilibrium For a system of forces to be in equilibrium: a) No resultant force must act. If a resultant force acts then the system would accelerate and would therefore not be in equilibrium. ∑ Fx = 0, ∑ Fy = 0 b) No resultant turning effect must act. ∑M = 0 Examples: 1) 8N 8N ∑ F = R = 8 – 8 = 0 ( state of equilibrium) 3 2) F3 = 2 N F1 = 4 N F2 = 4 N F4 = 2 N ∑ Fx = Rx = F1 – F2 = 4 – 4 = 0 N ∑ Fy = Ry = F3 – F4 = 2 – 2 = 0 N ( state of equilibrium ) 3) 8N 4N 4N 7N 6N 3N 6N 6N ∑ Fx = Rx = (4 + 6) - (7 + 3) = 0 N ∑ Fy = Ry = (4 + 8) - (6 + 6) = 0 N ( state of equilibrium ) 4 4) For three forces F1, F2 and F3 acting at point P, if ∑ F = 0, they will form a closed triangle. F1 F1 F3 F2 F2 F3 2.6 Vector Diagram Of Forces i) Definition: It is common for more than one force to act on a body simultaneously. When this occurs the forces are said to be concurrent, if they act at the same point, and coplanar if they lie within the same plane. ii) Vector Addition of Forces The net force that acts on an object when two or more forces act on it is known as the resultant force. If the forces act in the same straight line the resultant is found by simple addition or subtraction as shown in Figure 2.6.1. 5N 3N 4N a) R = 5 + 3 = 8 N 7N b) R = 7 - 3 = 8 N Figure 2.6.1 5 The resultant of forces that do not act in the same straight line can be determined by using the parallelogram law. The parallelogram law states that if two forces acting at a point are represented in size and direction by the sides of a parallelogram drawn from the point, their resultant is represented in size and direction by the diagonal of the parallogram drawn from the point. Using the parallelogram law, the resultant of the forces P and Q in Figure 2.6.2(a) is thus represented by R in Figure 2.6.2(b). P R P θ O O Q Q (a) (b) Figure 2.6.2 The resultant force, R can also be calculated by the cosine rule, ___________________ R = P2 + Q2 + 2PQ cos θ … [ because cos (180 – θ) = - cos θ ] R α P Q 180 O - θ The direction of the resultant R is given by the sine rule: sin α = _P sin θ R … because sin [ 180 – θ ] = sin θ 6 Examples: 1) Two concurrent forces of 3 N and 4 N act at a body at right angles to one another. Determine the magnitude and direction of the resultant force. Solution: (i) Scale 1 N : 1 cm 3 N = 3 cm, 4 N = 4 cm A C R 3 cm 900 O α 4 cm B Figure 2.6.3 From Figure 2.6.3, length of OC = 5 cm, therefore, R = 5 N and the direction of R, α = 37 0 ( ii ) P = 3 N, Q = 4 N, θ = 900 R = ( P2 + Q2 + 2PQcos θ ) = [ 32 + 42 + 2(3)(4)cos 900 ] = 5N P sin θ = ____ 3 sin 900 = 0.6 sin α = ____ R 5 α = sin-1 0.6 = 370 7 2) Compute the resultant forces produce by the following system of forces: i) ii) 1.5 kN 8N 800 600 O O 12 N 1 kN Solution: i) Scale 1 cm : 1 N A C R 8N 600 O α 12 N B Length of OC = 17.3 cm, R = 17.3 N Direction of R, α = 240 Or, R = ( P2 + Q2 + 2PQcos θ = [ 82 + 122 + 2(8)(12)cos 600 ] = 17.4 N sin α = ( 8 / 17.4 ) sin 600 = 0.398 α = sin-1 0.398 = 23.5 8 ii) A C Scale 1cm: 100 N 1.5 kN R 800 α O 1 kN B Length of OC = 19.3 cm, R = 19.3 x 100 N = 1930 N = 1.9 kN Direction of R, α = 500 Or, R = [ 1.52 + 12 + (1.5)(1)cos 800 ] = 1.9 kN sin α = (1.5/1.9) sin 800 = 0.777 α = sin-1 0.777 = 510 3) A resultant force of 130 N was produced by the action of two concurrent forces with the magnitudes of 70 N and 80 N. Find the magnitude of the angle made by the forces of 70 N and 80 N. Solution: 130 N 70 N θ 80 N 9 R2 = P2 + Q2 + 2PQcos θ cos θ = R2 - P2 - Q2 = 1302 - 702 - 802 = 0.5 2PQ 2(70)(80) θ = cos-1 0.5 = 600 2.7.1 Components of Forces Definition: Any single force may be split into two components. This is, in fact, the reverse of combining two forces into a resultant force. In this case the single force is resolved into two other forces. The most useful way to resolve a single force into two forces is to produce components which are perpendicular to each other. F F Fsinθ θ Fcosθ The force F can be resolved into the two perpendicular forces: Fcosθ and Fsinθ. 2.7.2 Resultant of a System of Forces Definition: When multiple forces act at a point, it is necessary to resolve each of the separate forces into a particular direction and then find the sum of all of the resolved components in that direction. For example, to find the resultant of the following four forces we need to resolve each force into the directions Ox and Oy and sum for each direction. We can then find the resultant of the two concurrent forces. 10 y 2N 4N O x 600 5 N 200 6N Resolving in the direction Ox ( ): Total force, ∑ Fx = Rx = 4 + 6cos 300 – 5cos 700 = 7.486 N Note that 2 N force is perpendicular to Ox and therefore has no component in the Ox direction. Resolving in the direction Oy (taking upwards as positive, ): Total force, ∑ Fy = Ry = 2 - 6cos 600 - 5cos 200 = - 5.70 N Note: The result is negative indicating that it is in the downward direction. The 4 N force has no component in the Oy direction. Once all four forces have been resolved into two perpendicular components, the resultant force and direction can be determined. 7.49 N θ 5.70 N R ______________ R = 7.492 + 5.702 = 9.41 N tan θ = 5.70 7.49 θ = tan-1 (5.70/7.49) = 370 11 Lami’s Theorem: The triangle of forces is used to analyse three force problems where all of the vector magnitudes are known or may be determined easily. In situations where the angles between the forces are known or may be easier to use Lami’s theorem, which is an adaptation of the sine rule. If three concurrent forces are in equilibrium, each force is proportional to the sine of the angle between the other two forces. F1 θ2 θ3 F3 θ1 F2 F1 F2 F3 -------- = -------- = --------sin θ1 sin θ2 sin θ3 Examples 1) Use Lami’s theorem to find the values of F1 and F2 F1 15 N γ 1300 1200 F2 12 Solution F1 15 ---------- = ---------sin 1200 sin 1300 F2 = 15 sin 1200 sin 1300 = 17 N γ = 3600 - 1300 – 1200 = 1100 F2 ---------- = sin 1100 F2 15 ----------sin 1300 15 sin 1100 sin 1300 = = 18.4 N 2) The diagram shows a particle in equilibrium being acted on by three forces. Find the value P. P 1200 900 Q 20 N . Solution P 20 --------- = ---------sin 900 sin 1200 P = 20 sin 900 sin 1200 = 23 N 13 3) A wooden block’s weight w is 20 N. It is suspended from the ceiling by two ropes with angles at 600 and 300. Find the tensions of the rope T1 and T2. 300 600 T1 T2 Solution T1 T2 900 1500 1200 20 N T1 20 ---------- = --------sin 1200 sin 900 T1 = 20 sin 120 sin 900 = 17.3 N T2 20 ----------- = ----------sin 1500 sin 900 T2 = 20 sin 1500 sin 900 = 10 N 14 4) In an experiment, a student hangs a 2 kg mass with a long string at a top end of a tripod. A short horizontal string attached to a spring balance is used to pull at the middle of the long string until it makes an angle of θ with the vertical as shown in figure below. If the reading of the spring balance is 30 N, find a) the angle θ that the string makes with the vertical, b) the tension T in the supporting string T θ 30 N Solution a) T cosθ θ 30 N Tsin θ 20 N ∑ Fx = 30 – T sin θ = 0 T sin θ = 30 ---- ( I ) ∑ Fy = T cos θ – 20 = 0 T cos θ = 20 ---- ( II ) 15 Dividing equation ( I ) by equation ( II ), T sin θ = 30 T cos θ 20 tan θ = 1.5 θ = 56.30 c) Replacing θ = 56.30 into equation ( I ), T sin 56.30 = 30 30 T = ----------- = 36 N sin 56.30 2.7 Definition of Force Moment If one or more non-concurrent forces act on a rigid body they may cause it to rotate about an axis. The turning effect of non-concurrent forces acting on a rigid body is measured by the moment of the force about this axis. The moment of a force about an axis is the product of the magnitude of the force and the perpendicular distance from the line of action of the force to the axis. x d O F The diagram shows a rigid bar being acted on by a force F at a pont x on the bar. The force causes the bar to rotate about a point O. The moment of the force about O is the product F.d. The moment of the force F about the point x is zero. The SI unit of the moment of a force is the newton-metre (Nm). Anti-clockwise moments of forces are usually taken to be positive and clockwise moments are usually taken to be negative. 16 Examples 1) Find the magnitude and sense of the moment of the given force about O? 8N O 2m MO = F x d = 8 x 2 = 16 Nm ( ) 2) Determine the net moment at O produced by the action of the two forces below. 2m O 3m 10 N 5N ∑ MO = ( 10 x 5 ) - ( 5 x 2 ) = 50 - 10 = 40 Nm 3) Find the magnitude and sense of the moment of the given force about O? 4m O 300 18 N 17 M0 = ( 18 sin 300 ) x 4 = 36 Nm 4) The following diagram shows a uniform square lamina ABCD subject to several forces. Find the resultant moment about A. D 6N E C 4N 2N 1N A 2m B 1m 3N MA = ( 6 x 2 ) - ( 4 x 1 ) - ( 1 x 2) = 6 Nm 5) In the diagram below, the rod XY is free to rotate about the point X. The forces shown are 1 m, 3 m and 7 m from X respectively. Find the moment about X of each of the forces and then determine the magnitude and sense of the resultant moment about X. 15 N X 300 5N Y 10 N Mx = ( 15 x 1 ) - ( 5 x 3 ) - [ ( 10sin 300 ) x 7 ] = - 35 Nm 2.7 Forces in Mechanical System Types of forces in mechanics include weight, pushes and pulls, friction, normal reaction, tension and thrust. - NOT FOR SALE - 18