Resposta halliday cap 12

Chapter 12

1. (a) The center of mass is given by

[pic]

(b) Similarly, we have

[pic]

(c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have

[pic]

(d) Similarly, ycog = [0 + (2.00)(m)(7.80) + (4.00)(m)(7.60) + (4.00)(m)(7.40) + (2.00)(m)(7.60) + 0]/(8.00m + 7.80m + 7.60m + 7.40m + 7.60m + 7.80m) = 1.97 m.

2. 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 [pic] is exerted, we find (since the acceleration is zero) 2T 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 – 2θ is the angle between the two segments, then we find α = 120º.

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 [pic] is exerted, we find (since the acceleration is zero) 2T sinθ = 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

[pic]

Therefore, T = F/(2sinθ ) = 7.92 × 103 N.

4. From [pic], 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.

5. Three forces act on the sphere: the tension force [pic] of the rope (acting along the rope), the force of the wall [pic] (acting