This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Bell, Evan Homework 8 Due: Mar 20 2005, 11:00 pm Inst: Furlan 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Two objects, of masses 30 kg and 54 kg, are hung from the ends of a stick that is 70 cm long and has marks every 10 cm, as shown. 30 kg 54 kg A B C D E F G 10 20 30 40 50 60 If the mass of the stick is negligible, at which of the points indicated should a cord be attached if the stick is to remain horizontal when suspended from the cord? 1. E 2. D 3. F correct 4. C 5. G 6. A 7. B Explanation: Let : = 70 cm , m 1 = 30 kg , m 2 = 54 kg . For static equilibrium, net = 0. Denote x the distance from the left end point of the stick to the point where the cord is attached. m 1 g x m 2 g [  x ] = = 0 ( m 1 m 2 ) x = m 2 x = m 2 m 1 m 2 = (54 kg)(70 cm) 30 kg 54 kg = 45 cm . Therefore the point should be point F . 002 (part 1 of 2) 10 points A uniform rod pivoted at one end point O is free to swing in a vertical plane in a gravita tional field. However, it is held in equilibrium by a force F at its other end. The rod makes an angle 36 with the hori zontal. The length of the rod is 7 m. The force makes and angle 58 with the horizontal. The weight of the rod is 7 . 4 N. x y 7 m F 5 8 3 6 O What is the magnitude of the force F ? Correct answer: 7 . 99068 N. Explanation: Let : = 7 m , W = 7 . 4 N , = 36 , and = 58 . Bell, Evan Homework 8 Due: Mar 20 2005, 11:00 pm Inst: Furlan 2 x y W F O R Basic Concepts: For static equilibrium, we have X i F x i = 0 , (1) X i F y i = 0 , and (2) X i i = 0 . (3) Solution: Using the torque equation, about point O , (Eq. 3 with counter clock wise positive), we have 2 W cos  F cos sin + F sin cos = 0 W 2 F sin + F tan cos = 0 , so F = W 2[sin  cos tan ] (4) = (7 . 4 N) 2[sin58  cos58 tan36 ] = 7 . 99068 N . Alternate Solution: The angle between the force F and the rod is =  = (58 ) (36 ) = 22 . Using the torque equation, about point O , (Eq. 3 with counter clockwise positive), we have 2 W cos = F sin , so F = W 2 cos sin (5) = (7 . 4 N) 2 cos36 sin22 = 7 . 99068 N . 003 (part 2 of 2) 10 points What is the magnitude of the force the pivot exerts on the rod at point O ? Correct answer: 4 . 28008 N. Explanation: Using the translational equations (Eqs. 1 and 2), where R x and R y are the horizontal and vertical components of the force R at the pivot, we have F x + R x = 0 , so (1) R x = F x = F cos = (7 . 99068 N)cos58 = 4 . 23442 N , and F y + R y = W , so (2) R y = W F y = W F sin = (7 . 4 N) (7 . 99068 N)sin58 = 0 . 623516 N , and k ~ R k = q R 2 x + R 2 y = q ( 4 . 23442 N) 2 + (0 . 623516 N) 2 = 4 . 28008 N ....
View
Full
Document
This note was uploaded on 03/30/2008 for the course PHYS 250 taught by Professor Gallis,michaelro during the Spring '07 term at Pennsylvania State University, University Park.
 Spring '07
 GALLIS,MICHAELRO
 Work

Click to edit the document details