ch12 - 1(a The center of mass is given by xcom =[0 0...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
1. (a) The center of mass is given by x com = [0 + 0 + 0 + ( m )(2.00) + ( m )(2.00) + ( m )(2.00)]/6.00 m = 1.00 m. (b) Similarly, y com = [0 + ( m )(2.00) + ( m )(4.00) + ( m )(4.00) + ( m )(2.00) + 0]/6 m = 2.00 m. (c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have x cog = x 1 m 1 g 1 + x 2 m 2 g 2 + x 3 m 3 g 3 + x 4 m 4 g 4 + x 5 m 5 g 5 + x 6 m 6 g 6 m 1 g 1 + m 2 g 2 + m 3 g 3 + m 4 g 4 + m 5 g 5 + m 6 g 6 = 0.987 m. (d) Similarly, y cog = [0 + (2.00)( m )(7.80) + (4.00)( m )(7.60) + (4.00)( m )(7.40) + (2.00)( m )(7.60) + 0]/(8.00 m + 7.80 m + 7.60 m + 7.40 m + 7.60 m + 7.80 m ) = 1.97 m.
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
2. From & & & τ = × r F , we note that persons 1 through 4 exert torques pointing out of the page (relative to the fulcrum), and persons 5 through 8 exert torques pointing into the page. (a) Among persons 1 through 4, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 2. (b) Among persons 5 through 8, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 7.
Image of page 2
3. 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). By analyzing the forces at the “kink” where & F is exerted, we find (since the acceleration is zero) 2 T sin θ = F , where θ is the angle (taken positive) between each segment of the string and its “relaxed” position (when the two segments are colinear). In this problem, we have 1 0.35m tan 11.5 . 1.72m θ § · = = ° ¨ ¸ © ¹ Therefore, T = F /(2sin θ )= 7.92 × 10 3 N.
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
4. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem). By analyzing the forces at the “kink” where & F is exerted, we find (since the acceleration is zero) 2 T sin θ = F , where θ 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 θ = 30º. Since α = 180º – 2 θ is the angle between the two segments, then we find α = 120º.
Image of page 4
5. 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 θ be the angle between the rope and the vertical. Then, the vertical component of Newton’s second law is T cos θ mg = 0. The horizontal component is F N T sin θ = 0. (a) We solve the first equation for the tension: T = mg / cos θ . We substitute cos θ = + L L r / 2 2 to obtain 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: sin N F T θ = . Using sin θ = + r L r / 2 2 , we obtain 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 + = = = = = + +
Image of page 5

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

View Full Document Right Arrow Icon
6. Let " 1 15 = . m and 2 (5.0 1.5) m 3.5 m = = " . We denote tension in the cable closer to the window as F 1 and that in the other cable as F 2 . The force of gravity on the scaffold itself (of magnitude m s g ) is at its midpoint, " 3 2 5 = . m from either end.
Image of page 6
Image of page 7
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