CHAPTER 5 FORCE, MASS AND ACCELERATION EXERCISE 83, Page 9. A car initially at rest accelerates uniformly to a speed of 55 km/h in 4 s. Determine the accelerating force required if the mass of the car is 800 kg. Initial velocity, v = 0 Final velocity, v = Time, t = 4 s Since v = v + at then acceleration, a = Hence, accelerating force, F = ma = km m h 55 000 = 5.78 m/s h km 3600s v v 5.78 0.09m / s = = t 800kg.09m / s = 873 N. The brakes are applied on the car in question when travelling at 55 km/h and it comes to rest uniformly in a distance of 50 m. Calculate the braking force and the time for the car to come to rest. Initial velocity, v = 55 km/h = = 5.78 m/s (from above) Final velocity, v = 0 Distance travelled, s = 50 m v = v + as from which, acceleration, a = v v 0 5.78 = =.33m / s s 50 Hence, braking force, F = ma = 800 kg.334 m/s = 867 N =.87 kn v v 5.78 0 Since v = v + at then time to come to rest, t = = = 6.55 s a.334 6
3. The tension in a rope lifting a crate vertically upwards is.8 kn. Determine its acceleration if the mass of the crate is 70 kg. T mg = ma i.e. 800 70 9.8 = 70 a from which, acceleration, a = 800 70 9.8 70 = 0.560 m/s 4. A ship is travelling at 8 km/h when it stops its engines. It drifts for a distance of 0.6 km and its speed is then 4 km/h. Determine the value of the forces opposing the motion of the ship, assuming the reduction in speed is uniform and the mass of the ship is 000 t. Initial velocity, v = 8 km/h = Final velocity, v = 4 3.6 km m h 8 8 000 = = 5 m/s h km 3600s 3.6 = 3.889 m/s Distance travelled, s = 0.6 km = 600 m v = v + as from which, acceleration, a = v v 3.889 5 = = - 8.3 0 m / s s 600 3 Hence, force opposing motion, F = ma = 000 000 kg 8.3 0 m / s 3 = 6459 N = 6.5 kn 5. A cage having a mass of t is being lowered down a mineshaft. It moves from rest with an acceleration of 4 m/s, until it is travelling at 5 m/s. It then travels at constant speed for 700 m and finally comes to rest in 6 s. Calculate the tension in the cable supporting the cage during (a) the initial period of acceleration, (b) the period of constant speed travel, (c) the final retardation period. 7
(a) Initial tension in cable, T = mg ma = m(g a) = 000(9.8 4) = 60 N =.6 kn (b) Tension in cable during constant speed, T = mg m a = mg 0 = 000 9.8 (c) Tension in retardation period, T 3 = mg m a 3 = 960 N = 9.6 kn where a 3 = v u 0 5 = =.5m / s t 6 Hence, tension, T 3 = mg m a 3 = 960 000(-.5) = 460 N = 4.6 kn 6. A miner having a mass of 80 kg is standing in the cage of problem 5. Determine the reaction force between the man and the floor of the cage during (a) the initial period of acceleration, (b) the period of constant speed travel, and (c) the final retardation period. (a) Reaction during initial acceleration = mg m a = m(g a ) = 80(9.8 4) = 464.8 N (b) Reaction during constant speed = mg m a = mg 0 = 80 9.8 = 784.8 N (c) Reaction in retardation period = mg m a 3 = 80(9.8 - -.5) = 80 (9.8 +.5) = 984.8 N 7. During an experiment, masses of 4 kg and 5 kg are attached to a thread and the thread is passed over a pulley so that both masses hang vertically downwards and are at the same height. When the system is released, find (a) the acceleration of the system, and (b) the tension in the thread, assuming no losses in the system. Let m = 4 kg and m = 5 kg 8
Consider the motion of mass m m g T = m a i.e. 5g T = 5a from which, a = 5g T g T = () 5 5 Consider the motion of mass m T - m g = m a T mg T 4g T from which, a = = = g () m 4 4 Equating equations () and () gives: T g = T g 5 4 i.e. g = T T 5T + 4T 9T + = = 4 5 0 0 i.e. 9T 0 = 9.8 and tension in thread, T = 0 9.8 9 = 43.6 N From (), acceleration, a = T 43.6 g = 9.8 =.09 5 5 m/s 9
EXERCISE 84, Page 93. Calculate the centripetal force acting on a vehicle of mass tonne when travelling round a bend of radius 5 m at 40 km/h. If this force should not exceed 750 N, determine the reduction in speed of the vehicle to meet this requirement. Centripetal acceleration = v r where v = km m h 40 000 =. m/s and r = 5 m h km 3600s Hence, centripetal acceleration, a =. 5 = 0.988 m/s Centripetal force = ma = 000 kg 0.988 m/s = 988 N If centripetal force 750 N ma then a = 750 N v = 0.75m / s = 000 kg r i.e. v = 0.75 5 and v = 0.75 5 = 9.68 m/s 9.68 m/s = m km 3600s 9.68 = 34.86 km/h s 000m h Hence the speed reduces form 40 km/h to 34.86 km/h. A speed-boat negotiates an S-bend consisting of two circular arcs of radii 00 m and 50 m. If the speed of the boat is constant at 34 km/h, determine the change in acceleration when leaving one arc and entering the other. 34 km/h = km 000 m h 34 = 9.444 m/s h km 3600s Acceleration, v 9.444 a = = = 0.89m / s r 00 30
Acceleration, a v 9.444 = = = 0.595m / s r 50 Change of acceleration = a a = 0.89 0.595 = 0.96 m/s i.e. change in acceleration = 0.3 m/s 3. An object is suspended by a thread 400 mm long and both object and thread move in a horizontal circle with a constant angular velocity of 3.0 rad/s. If the tension in the thread is 36 N, determine the mass of the object. Centripetal force (i.e. tension in thread) = mv r = 36 N The angular velocity, ω = 3.0 rad/s and radius, r = 400 mm = 0.4 m. Since linear velocity v = ωr, v = 3.0 0.4 =. m/s, and since F = mv r Fr, then m = v 36 0.4 i.e. mass of object, m = = 0 kg. 3
EXERCISE 85, Page 95. Calculate the mass moment of inertia of a thin rod, of length 0.5 m and mass 0. kg, about its centroid. Mass moment of inertia of a thin rod about its centroid, I = ml 0. 0.5 3 = = 4.67 0 kg m or 0.00467 kg m. Calculate the mass moment of inertia of the thin rod of Problem, about an end. Mass moment of inertia of a thin rod about an end, I = ml 0. 0.5 = = 0.0667 kg m 3 3 3. Calculate the mass moment of inertia of a solid disc of uniform thickness about its centroid. The diameter of the disc is 0.3 m and its thickness is 0.08 m. The density of its material of construction is 7860 kg/m 3. Mass moment of inertia of a solid disc of uniform thickness about its centroid, I = R 3 0.5 ρπ R t = 7860kg / m π 0.5 0.08 = 0.50 kg m 4. If a hole of diameter 0. m is drilled through the centre of the disc of Problem 3, what will be its mass moment of inertia about its centroid? Mass moment of inertia of a solid disc of uniform thickness about its centroid, I = ( ) ( ) R + R 3 (0.5 + 0. ) ρπ R R t = 7860 kg / m π (0.5 0. ) 0.08 = 0.40 kg m 3
EXERCISE 86, Page 95 Answers found from within the text of the chapter, pages 88 to 94. EXERCISE 87, Page 95. (c). (b) 3. (a) 4. (d) 5. (a) 6. (b) 7. (b) 8. (a) 9. (a) 0. (d). (d). (c) 3. (b) 33