Dynamic equilibrium: object moves with constant velocity in a straight line. We note that F net a s are both vector quantities, so in terms of their components, (F net ) x = i (F i ) x = 0, a x = i (a i ) x = 0, (4.11) with similar equations for the y- z-components. Example: (Static equilibrium) 26 the three ropes in figure 36 are tied to a small, very light ring. Two of the ropes are anchored to the walls at right angles, the third rope pulls as shown. What are T 1 T 2, the magnitudes of the tension forces in the first two ropes? Figure 36: Static equilibrium. Solution: for the ring to be in static equilibrium, we must have, from Newton s First Law, equation (4.11), 26 Knight, Exercise 1, page 146 56
(F net ) x = (T 1 ) x + (T 2 ) x + (T 3 ) x = 0 (4.12) (F net ) x = (T 1 ) y + (T 2 ) y + (T 3 ) y = 0. (4.13) The components of the force are: x : (T 1 ) x = T 1, (T 2 ) y = 0, (T 3 ) x = T 3 cos 30 y : (T 1 ) y = 0, (T 2 ) y = T 2, (T 3 ) y = T 3 sin 30 Hence, using equations (4.12) (4.13), we have T 1 + 0 + T 3 cos 30 = 0 T 1 = 100 cos 30 = 50 3 N, (4.14) 0 + T 2 T 3 sin 30 = 0 T 2 = 100 sin 30 = 50 N, (4.15) respectively, where we have used T 3 = 100 N. For an example concerning dynamic equilibrium, see Additional Materials. 27 4.8 Non-equilibrium Conditions The equilibrium conditions (4.9) (4.10) are a special case of the more general non-equilibrium situation where 27 Equilibrium Particle Dynamics Example, http://www.nanotech.uwaterloo.ca/ ne131/ 57
The net force on an object is non-zero. The object accelerates is no longer in equilibrium. It is Newton s 2nd Law which therefore provides the more general link between force motion: we write F net = i F i = ma s, (4.16) with the corresponding component equations being (F net ) x = i (F i ) x = ma x (4.17) (F net ) y = i (F i ) y = ma y. (4.18) Example: 28 in each of the diagrams in figure 37, 29 the forces are acting on a 2.0 kg object. For each figure, find the values a x a y, the x- y-components of the acceleration. Solution: considering each diagram in turn: Diagram 1: the x- y-components are as follows; x : (F 1 ) x = 4 N, (F 2 ) x = 0 N, (F 3 ) x = 2 N y : (F 1 ) y = 3 N, (F 2 ) y = 3 N, (F 3 ) y = 0 N. Thus, substituting in equations (4.17) (4.18) gives us 4 N 0 N 2 N = 2 a x a x = 1 m s 2, 28 Knight, Exercise 5, page 146 29 Knight, Figures 5.5(a),(b), page 146 58
Diagram 1 Diagram 2 Figure 37: Non-equilibrium motion. respectively. 3 N 3 N + 0 N = 2 a y a y = 0 m s 2, Diagram 2: the x- y-components are: x : (F 1 ) x = 0 N, (F 2 ) x = 4 N, (F 3 ) x = 0 N, (F 4 ) x = 0 N, (F 5 ) x = 2 N y : (F 1 ) y = 3 N, (F 2 ) y = 0 N, (F 3 ) y = 2 N, (F 4 ) y = 1 N, (F 5 ) y = 0 N. Substitution in equations (4.17) (4.18) gives 0 N + 4 N + 0 N + 0 N 2 N = 2 a x a x = 1 m s 2 3 N + 0 N 2 N 1 N + 0 N = 2 a y a y = 0 m s 2, respectively. 59
4.9 Common Types of Forces Revisited Let us apply Newton s laws to the forces discussed previously. 4.9.1 Weight Recall that mass weight are not the same thing: Mass is an intrinsic property of an object, indicating the amount of matter it contains. Weight is the force on an object due to its presence in the gravitational field of the Earth. From equation (4.2), they are related via the expression W = m a free fall, (4.19) where W has the magnitude W = mg, acts vertically downwards. Equation (4.19) is a form of Newton s 2nd Law; omitting the drag force due to air resistance, an object in free fall experiences a net force F net = W = m a free fall. (4.20) Apparent Weight We sense gravity indirectly through the normal force due to the contact surface The weight force presses down on the ground, chair, etc. The ground, chair, etc. exerts a normal force upwards upon us. Sting at rest on the ground (figure 38), we experience a net force of zero vertically downwards: Equilibrium: applying Newton s 1st Law, equation (4.11), (F net ) y = N W = 0, 60
i.e. N = W = m a free fall = m g. (4.21) Figure 38: Object at rest. i.e. in our inertial reference frame, we sense our weight as the normal force pushing up on our feet with a magnitude mg. What happens when there is a resulting vertical force acting upon us? For example, consider motion in an elevator Upwards acceleration: for an acceleration a s 0 upwards (figure 39): Non-equilibrium: applying New- ton s 2nd Law, equation (4.18): i.e. (F net ) y = N W = m a s, N = W + ma s. (4.22) Figure 39: Upwards acceleration in an elevator. In our non-inertial reference frame accelerating upwards, the normal force pushing upwards on our feet - our sensation of weight - has increased! Note: it is important to realise that 61
Our apparent weight W app has increased. Our true weight W is still the same. Substituting for our true weight in equation (4.22) defines the magnitude of our apparent weight to be the magnitude of the contact force N, i.e. ( W app = mg 1 + a ) ( s = W 1 + a ) s g g (upwards acceleration). (4.23) So this explains why we feel heavier when an elevator suddenly accelerates upwards. Downwards acceleration: for an acceleration a s a s < g (figure 40): 0 downwards, where Non-equilibrium: applying Newton s 2nd Law (4.18): i.e. (F net ) y = W N = m a s, N = W ma s. (4.24) Figure 40: Downwards acceleration in an elevator. In our non-inertial reference frame accelerating downwards, the normal force pushing upwards on our feet - our sensation of weight - has decreased! So by accelerating downwards in an elevator, we feel lighter than usual. By analogy with equation (4.23), our apparent weight is 62
( W app = mg 1 a ) ( s = W 1 a ) s g g (downwards acceleration, a s < g). (4.25) Apparent weight has several important applications: e.g. the upwards acceleration during a rocket launch is a s g under such conditions, equation (4.23) shows us that astronauts will experience an extremely large apparent weight. Weightlessness In the elevator example, what happens if the cable snaps? We will then be in free fall, i.e. ma s = mg, from equation (4.25), our apparent weight will be W app = 0. (4.26) We lose our sensation of weight there would be no normal force pushing on our feet thus be weightless. Note: although our apparent weight would be zero, our true weight is still of magnitude mg. 4.9.2 Friction Recall our classification of static, kinetic rolling friction. Static friction f s : the frictional force on an object which prevents it from slipping. Suppose we push an object with a horizontal force F push (figure 41 30 ) The box remains at rest, so it is in static equilibrium; from Newton s 1st Law, we have 30 Knight, Figures 5.9 (a),(b), page 132 63
Example of static friction. Force diagram for static friction. Figure 41: Static friction. f s = F push. (4.27) If F push is large enough, the object moves - this suggests the upper limit f s max to f s : f s < f s max object remains at rest. f s = f s max object slips. f s > f s max unphysical. Furthermore, f s max is proportional to the magnitude of the normal force, i.e. f s max N, or f s max = µ s N, (4.28) where we have introduced the coefficient of static friction, µ s, a number which depends upon the materials making up the object surface. Note: in equation (4.28), we are saying that f s max = µ s N, not f s = µ s N. Kinetic friction f k : the frictional force that appears when an object moves across the surface. When the object begins to slide (figure 42 31 ), 31 Knight, Figures 5.11 (a),(b), page 133 64