1. (a) The center of mass is given by
(b) Similarly, we have
(c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have
(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) = 1.97m.
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 = Ftherefore 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
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 intothe 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] (actinghorizontally away from the wall), and the force of gravity [pic] (acting downward). Since the sphere is in equilibrium they sum to zero. Let θ be the angle between the rope and the vertical. Then Newton’s second law gives
vertical component : T cos θ – mg = 0
horizontal component: FN – T sin θ = 0.
(a) We solve the first equation for the tension: T = mg/ cos θ. Wesubstitute [pic]to obtain
(b) We solve the second equation for the normal force: [pic]. Using[pic], we obtain
6. Our notation is as follows: M = 1360 kg is the mass of the automobile; L = 3.05 m is the horizontal distance between the axles; [pic]is the horizontal distance from the rear axle to the center of mass; F1 is the force exerted on each front wheel; and, F2 is the forceexerted on each back wheel.
(a) Taking torques about the rear axle, we find
(b) Equilibrium of forces leads to[pic]from which we obtain[pic].
7. We take the force of the left pedestal to be F1 at x = 0, where the x axis is along the diving board. We take the force of the right pedestal to be F2 and denote its position as x = d. W is the weight of the diver, located at x = L. Thefollowing two equations result from setting the sum of forces equal to zero (with upwards positive), and the sum of torques (about x2) equal to zero:
(a) The second equation gives
which should be rounded off to [pic]. Thus, [pic]
(b) Since F1 is negative, indicating that this force is downward.
(c) The first equation gives [pic]
which should be rounded off to [pic].Thus, [pic]
(d) The result is positive, indicating that this force is upward.
(e) The force of the diving board on the left pedestal is upward (opposite to the force of the pedestal on the diving board), so this pedestal is being stretched.
(f) The force of the diving board on the right pedestal is downward, so this pedestal is being compressed.
8. Let [pic]and[pic]. We denote tension...