05_InstSolManual_PDF_Part13 - same magnitude The box has...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
(b) Since the speed is decreasing, the acceleration is opposite to the direction of the motion and is to the right, as in part (a). The same equation applies as in part (a) and The accelerometer deflects to the left of vertical. (c) Constant speed means The hanging mass has no acceleration and hence no horizontal force on it. The accelerometer hangs vertically. 5.27. Set Up: Use coordinates where is parallel to the slope and directed up the slope. Let the skier have mass m . The normal force n exerted by the ground is perpendicular to the slope. Solve: (a) The free-body diagram for the skier is given in Figure 5.27. The weight mg of the skier has been replaced by its x and y components. gives and Her accelera- tion is directed down the slope. Figure 5.27 (b) and (c) Nothing changes in the free-body diagram and her acceleration is the same as in (a). 5.28. Set Up: In each free-body diagram take a positive coordinate direction to be the direction of the acceleration of the object. The box accelerates to the right and the bucket accelerates downward. Their accelerations have the
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: same magnitude. The box has mass Solve: (a) The free-body diagram for the box and bucket are shown in Figure 5.28. Figure 5.28 (b) applied to the box gives applied to the bucket gives Combining these two equations to eliminate T gives (c) gives v y 5 " 2 a y 1 y 2 y 2 5 " 2 1 3.58 m / s 2 21 1.50 m 2 5 3.28 m / s v y 2 5 v y 2 1 2 a y 1 y 2 y 2 y 2 y 5 1.50 m. a y 5 3.58 m / s 2 , v y 5 0, a 5 1 m bucket m box 1 m bucket 2 g 5 3.58 m / s 2 m bucket g 2 T 5 m bucket a . g F y 5 ma y T 5 m box a . g F x 5 ma x x y w box n T box x y m bucket g a T bucket a m box 5 1 375 N 2 / 1 9.80 m / s 2 2 5 38.3 kg. y n x mg cos24° mg sin24° 24° 4.0 m / s 2 , a x 5 2 g sin 24° 5 2 4.0 m / s 2 . 2 mg sin 24° 5 ma x g F x 5 ma x 1 x a 5 0. 25° b 5 25°. tan b 5 a g 5 4.5 m / s 2 9.80 m / s 2 . Applications of Newton’s Laws 5-13...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern