{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch12 - Chapter 12 1(a The center of mass is given by xcom 0...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
519 Chapter 12 1. (a) The center of mass is given by com 0 0 0 ( )(2.00 m) ( )(2.00 m) ( )(2.00 m) 1.00 m. 6 m m m x m (b) Similarly, we have com 0 ( )(2.00 m) ( )(4.00 m) ( )(4.00 m) ( )(2.00 m) 0 2.00 m. 6 m m m m y m (c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have 6 1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 cog 6 1 1 2 2 3 3 4 4 5 5 6 6 1 0.987 m. i i i i i i i x m g x m g x m g x m g x m g x m g x m g x m g m g m g m g m g m g m g (d) Similarly, we have 6 1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 cog 6 1 1 2 2 3 3 4 4 5 5 6 6 1 0 (2.00)(7.80 ) (4.00)(7.60 ) (4.00)(7.40 ) (2.00)(7.60 ) 0 8.0 7.80 7.60 7.40 7.60 7.80 1.97 m. i i i i i i i y m g y m g y m g y m g y m g y m g y m g y m g m g m g m g m g m g m g m m m m m m m m m m 2. Our notation is as follows: M = 1360 kg is the mass of the automobile; L = 3.35 m is the horizontal distance between the axles; (3.35 1.85) m 1.50 m is the horizontal distance from the rear axle to the center of mass; F 1 is the force exerted on each front wheel; and F 2 is the force exerted on each back wheel. (a) Taking torques about the rear axle, we find 2 3 1 (1360kg)(9.80m/s )(1.50m) 2.98 10 N. 2 2(3.35m) Mg F L
Image of page 1

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

View Full Document Right Arrow Icon
CHAPTER 12 520 (b) Equilibrium of forces leads to 1 2 2 2 , F F Mg from which we obtain 3 2 3.68 10 N F . 3. Three forces act on the sphere: the tension force T of the rope (acting along the rope), the force of the wall N F (acting horizontally away from the wall), and the force of gravity mg (acting downward). Since the sphere is in equilibrium they sum to zero. Let be the angle between the rope and the vertical. Then Newton’s second law gives vertical component : T cos mg = 0 horizontal component: F N T sin = 0. (a) We solve the first equation for the tension and obtain T = mg / cos . We then substitute cos L L r / 2 2 : 2 2 2 2 2 (0.85 kg)(9.8 m/s ) (0.080 m) (0.042 m) 9.4 N 0.080 m mg L r T L . (b) We solve the second equation for the normal force and obtain sin N F T . Using sin r L r / 2 2 , we have 2 2 2 2 2 2 2 (0.85 kg)(9.8 m/s )(0.042 m) 4.4 N. (0.080 m) N Tr mg L r r mgr F L L L r L r 4. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem in the text). By analyzing the forces at the “kink” where F is exerted, we find (since the acceleration is zero) 2 T sin = F , where is the angle (taken positive) between each segment of the string and its “relaxed” position (when the two segments are collinear). Setting T = F therefore yields = 30º. Since = 180º – 2 is the angle between the two segments, then we find = 120º.
Image of page 2
521 5. The object exerts a downward force of magnitude F = 3160 N at the midpoint of the rope, causing a “kink” similar to that shown for problem 10 (see the figure that accompanies that problem in the text). By analyzing the forces at the “kink” where F is
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the 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