Phys 141_HW Sol_Ch5_W08

Phys 141_HW Sol_Ch5_W08 - Cal Poly State University SLO...

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Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 Homework Solutions Chapter 5 Assigned Problems: 1, 3, 6, 8, 10, 11, 17, 31, 35, 39, 49, 52 5.1. 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 ( 29 net 1 2 3 0 N x x x x x F F T T T = Σ = + + = ( 29 net 1 2 3 0 N y y y y y F F T T T = Σ = + + = Evaluating the components of the force vectors from the free-body diagram: 1 1 x T T = - T 2 x = 0 N 3 3 cos30 x T T = ° T 1 y = 0 N 2 2 y T T = 3 3 sin30 y T T = - ° Using Newton’s first law: 1 3 cos30 0 N T T - + ° = 2 3 sin30 0 N T T - ° = Rearranging: ( 29 ( 29 1 3 cos30 100 N 0.8666 86.7 N T T = ° = = ( 29 ( 29 2 3 sin30 100 N 0.5 50.0 N T T = ° = = Assess: Since 3 T r acts closer to the x -axis than to the y -axis, it makes sense that 1 2 T T . 5.3. 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 θ as follows: 1 2 m sin 0.677 sin 0.667 41.8 3 m θ θ - = = = = ° Newton’s first law for this situation is
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Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 ( 29 net 1 2 1 2 0 N cos cos 0 N x x x x F F T T T T θ θ = Σ = + = ⇒ - + = ( 29 net 1 2 1 2 0 N sin sin 0 N y y y y y F F T T w T T w θ θ = Σ = + + = + - = The x -component equation means 1 2 T T = . From the y- component equation: 1 2 sin T w θ = ( 29 ( 29 2 1 20 kg 9.8 m/s 196 N 147 N 2sin 2sin 2sin41.8 1.333 w mg T θ θ = = = = = ° Assess: It’s to be expected that the two tensions are equal, since the speaker is suspended symmetrically from the two cables. That the two tensions add to considerably more than the weight of the speaker reflects the relatively large angle of suspension. 5.6. Visualize: Please refer to Figure Ex5.6. 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, ( 29 ( 29 net 2 5 N 1 N 3sin20 N 1.49 m/s 2 kg x x F a m - - ° = = = ( 29 ( 29 net 2 2.82 N 3cos20 N 0 m/s 2 kg y y F a m - ° = = = 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 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 net 2 net 2 2cos15 N 2sin15 N 3 N 0.28 m/s 2 kg 1.414 N 2sin15 N 2cos15 N 0 m/s 2 kg x x y y F a m F a m ° + ° - = = = - + ° - ° = = = 5.8.
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