Systes of Masses. Ignoring friction, calculate the acceleration of the syste below and the tension in the rope. Drawing individual free body diagras we get 4.0kg 7.0kg g 9.80 / s a?? g and g (4.0)(9.80) 39.N g g (7.0)(9.80) 68.6N or the 4.0 kg ass, the only force parallel to the acceleration (which is horizontal) is the tension force in the rope; the forces and g are perpendicular to the acceleration and should not be included. or the 7.0 kg ass, both of the forces and g are parallel to the acceleration (which is vertical). N When looking at the forces causing the syste to accelerate, tension is not included since this is an internal force. he only reaining force is therefore g. his would be the only horizontal force (or the only force parallel to the otion) in our linearized free body diagra for the whole syste: RRHS Physics
Using clockwise (or toward the 7.0 kg ass) as positive, a a t a t g g.0a 68.6 a 6. / s o find the tension it is necessary to set up an equation for one of the asses individually. If we do this for the 4.0 kg ass, we get a a (4.0)(6.) 5N. A.00 kg ass and a 3.00 kg ass are attached to a lightweight cord that passes over a pulley. he hanging asses are free to ove. (a) What is its acceleration? RRHS Physics
.00kg 3.00kg g 9.80 / s a? g g (.00)(9.80) 9.6N g g (3.00)(9.80) 9.4N reating the syste as one object and using the 3.00 kg ass as the positive end (clockwise) a a t g g a t g g 5.00a 9.4 9.6 a.96 / s (b) What is the tension force acting in the cord? o find the tension it is necessary to look at just one of the objects. Looking at the 3.00 kg object, a g g 3.00(.96) 9.4 a a 3.5N 3. Do Question # question again if the coefficient of friction between the top box and the table is 0.3. RRHS Physics 3
4.0kg 7.0kg 0.3 a?? g g (4.0)(9.80) 39.N We also need the force of friction on : f N g (0.3)(39.).N g g (7.0)(9.80) 68.6N Using the 7.0 kg ass end of the rope as positive, a a t g f a t g f.0a 68.6. a 5. / s o find the tension, we need to look at just one object. Looking at the 4.0 kg object, a f f 4.0(5.). a a 33N 4. What coefficient of static friction is necessary to prevent otion in the diagra below? 4 RRHS Physics
3.0kg g g g g 5.0kg (3.0)(9.80) (5.0)(9.80) s? 9.4N 49N a 0 Since there is no otion, the acceleration ust be zero. a t g f 0 g f g f g s N g s g 49 (9.4) s.7 s a 5. What coefficient of kinetic friction is needed in the last question for a constant speed? 3.0kg 5.0kg? k a 0 Since the acceleration is again zero, the sae forces have to balance as in the previous question, so the ath is the exact sae as the previous question. he only RRHS Physics 5
difference is that this tie we are finding the coefficient of kinetic friction since the syste is in otion. k.7 6. ind the acceleration of the following syste (a) if there is no friction..0kg g g g g g g 5.0kg (.0)(9.80) (5.0)(9.80) (6.0)(9.80) 3 6.0kg 9.6N 49N 58.8N a? Since all of the tension forces are internal forces, the only forces that contribute to the otion are g and g. We will use the 6.0 kg end of the rope as the positive direction: a a t g g a t g g 3.0a 9.6 58.8 a 3.0 / s (b) if the coefficient of friction between the box and the table is 0.. 6 RRHS Physics
.0kg 3 5.0kg 6.0kg 0. a? g g (.0)(9.80) 9.6N g g (5.0)(9.80) 49N g g (6.0)(9.80) 58.8N f N g (0.)(49) 5.9N Since all of the tension forces are internal forces, the only forces that contribute to the otion are g, f and g. We will use the 6.0 kg end of the rope as the positive direction: a a t g f g3 a t g f g3 3.0a 9.6 5.9 58.8 a.6 / s 7. A.5 kg counterweight is connected (over a pulley) to a 4.5 kg window that slides vertically in its frae. How uch force ust you exert to start the window opening with an upward acceleration of 0.8 /s? RRHS Physics 7
4.5kg.5kg g 9.80 / s a 0.8 / s p? g g (4.5)(9.80) 44.N g g (.5)(9.80) 4.5N Looking at the syste as one object (so we will ignore the internal tension forces) and using the counterweight end of the rope as the positive end, a t g g p t g g p 7.0(0.8) 44. 4.5 p a a N p 8. A. kg ass and a 3. kg ass are attached by a assless rope hanging over a pulley. he two asses are initially.60 above the ground and the pulley is 4.0 above the ground. What axiu height does the. kg object reach after the syste is released fro rest? 8 RRHS Physics
he 3. kg ass will fall.6 to the floor; during this tie, the. kg ass will be accelerating upward. We ust find this acceleration so that we can deterine the speed of the. kg ass when it stops accelerating upward (at a height of 3. fro the floor)..kg 3.kg g 9.80 / s a? g g (.)(9.80).8N g g (3.)(9.80) 3.4N reating the syste as one object and using the 3. kg ass as the positive end (clockwise) a a t g g a t g g 4.4a 3.4.8 a 4.45 / s We can now find the upward velocity of the. kg ass when it is at a height of 3.: v 0 v i f? a 4.45 / s d.6 vf vi d a v f 0.6 (4.45) v f 3.77 / s We can now calculate how uch higher it goes before gravity brings it to a stop: RRHS Physics 9
v 3.77 / s v i f 0 a 9.80 / s d? So the. kg ass will be d vf vi a 0 3.77 ( 9.80) 0.73.6.6 0.7 3.9 above the floor (and still below the 4.0 high pulley) 0 RRHS Physics