Phys 141_HW Sol_Ch5_W08

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern