6 1 Model We can assume that the ring is a single massless particle in static equilibrium Visualize Solve Written in component form Newton s first law is F x net F 0 N T 2 T 3 T 1 x x x x net yF F y T 1 y T 2 y T 3 y 0 N Evaluating the components of the force vectors from the free body diagram T 1 x T T 1 2 x 0 N xT 3 T 3 cos30 yT 1 0 N yT 2 T 2 yT 3 T 3 sin 30 T T 1 3 cos30 0 N T 2 T 3 sin 30 0 N T 3 cos30 100 N 0 8666 86 7 N T 2 T 3 sin 30 100 N 0 5 50 0 N acts closer to the x axis than to the y axis it makes sense that 1 T 2 T Using Newton s first law Rearranging T 1 cid 71 3T Assess Since 6 2 Model We can assume that the ring is a particle Visualize This is a static equilibrium problem We will ignore the weight of the ring because it is very light so the only three forces are the tension forces shown in the free body diagram Note that the diagram defines the angle Solve Because the ring is in equilibrium it must obey and y components This is a vector equation so it has both x cid 71 F net 0 N net F x net F y T 3 cos T 2 0 N T 3 cos T 2 T T 1 3 sin 0 N T 3 sin T 1 We have two equations in the two unknowns 3T and Divide the y equation by the x equation T 3 T 3 sin cos tan T 1 T 2 80 N 50 N 1 6 tan 1 1 6 58 Now we can use the x equation to find T 3 T 2 cos 50 N cos58 94 N The tension in the third rope is 94 N directed 58 below the horizontal 6 3 Model We assume the speaker is a particle in static equilibrium under the influence of three forces gravity and the tensions in the two cables Visualize Solve From the lengths of the cables and the distance below the ceiling we can calculate as follows sin 0 677 1 sin 0 667 41 8 2 m 3 m Newton s first law for this situation is net F x F x T 1 x T 2 x 0 N T 1 cos T 2 cos 0 N net F y F y T 1 y T 2 y w y 0 N T 1 sin T 2 sin w 0 N T The x component equation means 1 T 2 From the y component equation 12 sinT w T 1 w 2sin mg 2sin 20 kg 9 8 m s 2sin 41 8 2 196 N 1 333 147 N It s to be expected that the two tensions are equal since the speaker is suspended symmetrically from Assess the two cables That the two tensions add to considerably more than the weight of the speaker reflects the relatively large angle of suspension 6 4 Model We can assume that the coach and his sled are a particle being towed at a constant velocity by the two ropes with friction providing the force that resists the pullers Visualize Solve Since the sled is not accelerating it is in dynamic equilibrium and Newton s first law applies net F x F x T 1 x T 2 x f x k 0 N net F y F y T 1 y T 2 y f k y 0 N From the free body diagram T 1 cos T 2 cos f k 0 N T 1 sin 1 2 1 2 1 T 2 2 sin 1 2 0 N 0 N T From the second of these equations 1 T 2 Then from the first T 12 cos10 1000 N T 1 1000 N 1000 N 1 970 2cos10 508 N Assess The two tensions are equal as expected since the two players are pulling at the same angle The two add up to only slightly more than 1000 N which makes sense because the angle at which the two players are pulling is small 6 5 Visualize Please refer to the Figure EX6 5 Solve Applying Newton s second law to the diagram on the left a x net F m x 4 N 2 N 2 kg 1 0 m s 2 a y net F m y 3 N 3 N 2 kg 0 m s 2 For the diagram on the right a x net F m x 4 N 2 N 2 kg 1 0 m s 2 a y net F m y 3 N 1 N 2 N 2 kg 0 m s 2 6 6 Visualize Please refer to Figure EX6 6 Solve For the diagram on the left three of the vectors lie along the axes of the tilted coordinate system Notice that the angle between the 3 N force and the y axis is the same 20 by which the coordinates are tilted Applying Newton s second law For the diagram on the right the 2 newton force in the first quadrant makes an angle of 15 with the positive x axis The other 2 newton force makes an angle of 15 with the negative y axis The accelerations are a x F x net m 5 N 1 N 3sin 20 N 2 kg 1 49 m s 2 a y F net m 2 82 N 3cos 20 N 2 kg 0 m s 2 y a x a y x y F F net m net m 2cos15 N 2sin15 N 3 N 2 kg 0 28 m s 2 1 414 N 2sin15 N 2cos15 N 2 kg 0 m s 2 6 7 Visualize Please refer to Figure EX6 7 Solve a Apply Newton s second law in both the x and y directions F net x 5 0 N cos37 2 0 N 5 0 kg a x yF net 2 0 N 5 0 N sin 37 5 0 N 5 0 kg a y xa 0 40 m s 2 ya 0 0 m s 2 F net x 3 0 N 5 0 N sin 37 2 0 N 5 0 kg a x xa 0 80 m s 2 ya 0 0 m s 2 F net y 4 0 N 5 0 N cos37 5 0 kg a y b The angle that the 5 0 N force makes with the y axis is 37 Apply Newton s second law for both the x and y direction Assess The orientation of the coordinate axes is chosen for convenience and does not always need to conform to the horizontal and vertical 6 8 Visualize Please refer to Figure EX6 8 Solve We can use the constant slopes of the three segments of the …
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