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Unformatted text preview: 11.1: Take the origin to be at the center of the small ball; then, m 387 . kg 00 . 3 ) m 580 . )( kg 00 . 2 ( kg)(0) 00 . 1 ( cm = + = x from the center of the small ball. 11.2: The calculation of Exercise 11.1 becomes m 351 . kg 50 . 4 ) m 580 . )( kg 00 . 2 ( ) m 280 . )( kg 50 . 1 ( ) )( kg 00 . 1 ( cm = + + = x This result is smaller than the one obtained in Exercise 11.1. 11.3: In the notation of Example 11.1, take the origin to be the point , S and let the child’s distance from this point be . x Then, m, 125 . 1 2 , ) 2 ( cm = = = + +- = m MD x m M mx D M s which is , 2 ) 2 2 ( D L- halfway between the point S and the end of the plank. 11.4: a) The force is applied at the center of mass, so the applied force must have the same magnitude as the weight of the door, or N. 300 In this case, the hinge exerts no force. b) With respect to the hinge, the moment arm of the applied force is twice the distance to the center of mass, so the force has half the magnitude of the weight, or N 150 . The hinge supplies an upward force of N. 150 N 150 N 300 =- 11.5: kN, 45 . 5 so ), m . 10 )( N 2800 ( 40 sin ) m . 8 ( = = ° F F keeping an extra figure. 11.6: The other person lifts with a force of N. 100 N 60 N 160 =- Taking torques about the point where the N- 60 force is applied, m. 40 . 2 N 100 N 160 ) m 50 . 1 ( or ), m 50 . 1 )( N 160 ( ) N 100 ( = = = x x 11.7: If the board is taken to be massless, the weight of the motor is the sum of the applied forces, N. 1000 The motor is a distance m 200 . 1 ) N 1000 ( ) N 600 )( m 00 . 2 ( = from the end where the 400-N force is applied. 11.8: The weight of the motor is N. 800 N 200 N 600 N 400 =- + Of the myriad ways to do this problem, a sneaky way is to say that the lifters each exert N 100 to the lift the board, leaving N 500 and N 300 to the lift the motor. Then, the distance of the motor from the end where the 600-N force is applied is m 75 . ) N 800 ( ) N 300 )( m 00 . 2 ( = .The center of gravity is located at m 80 . ) N 1000 ( ) m 75 . )( N 800 ( ) m . 1 )( N 200 ( = + from the end where the N 600 force is applied. 11.9: The torque due to , cot is h θ h T T D Lw x x- =- and the torque due to Lw D T T y y = is . The sum of these torques is ). cot 1 ( θ Lw D h- From Figure (11.9(b)), , tan θ D h = so the net torque due to the tension in the tendon is zero. 11.10: a) Since the wall is frictionless, the only vertical forces are the weights of the man and the ladder, and the normal force. For the vertical forces to balance, N, 900 N 740 N 160 m 1 2 = + = + = w w n and the maximum frictional forces is N 360 ) N 900 )( 40 . ( 2 s = = n μ (see Figure 11.7(b)). b) Note that the ladder makes contact with the wall at a height of 4.0 m above the ground. Balancing torques about the point of contact with the ground, m, N 684 ) N 740 ))( 5 3 )( m . 1 ( ) N 160 )( m 5 . 1 ( ) m . 4 ( 1 ⋅ = + = n so N, . 171 1 = n keeping extra figures. This horizontal force about must be balanced by keeping extra figures....
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This note was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.
- Fall '07