This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: P207 - Fall 2007 Solutions Assignment 10 1. Chapter 10, Problem 48 O = F A R A sin A + F B R B sin B + F C R C sin C = (10 N)(8 m)sin(135 )- (16 N)(4 m)sin(90 ) + (19 N)(3 m)sin(160 ) = 12 Nm (Counter Clockwise) 2. Chapter 12, Question 5. (a) The vertical component of the tension of the cord must be the same in all three cases so that the torque about the hinge will be zero. So (2) has the minimum tension. (1) and (3) have the same tension. (b) The torque about the far end of the rod is due only to the vertical force of the wall on the hinge. So it must be the same for all three configurations. (c) The net horizontal force is zero. The horizontal force from the hinge must balance the horizontal force of the tension of the cord. The horizontal force from the hinge will be equal magnitude but opposite direction for configurations (1) and (3). And zero for (2). 3. Chapter 12, Problem 9 (a) The torque about the base of the ladder is zero. = W c r c sin + W l r cm,l sin - F w r l sin( / 2- ) = 0 where W c and r c are the weight of the cleaner and his distance from the base of the ladder, W l and r cm,l are the weight of the ladder and the distance of the center of mass from the base, and F w and r l are the force of the window on the ladder and the length of the ladder. is the angle of the ladder with respect to the vertical. = sin- 1 (2 . 5 / 5) = 30 . Note that the weights of cleaner and ladder are in the vertical direction. The force of the window on the ladder is in the horizontal direction. Solve for F w = 1 r l sin( / 2- ) ( m c gr c sin + m l gr cm,l sin ) = 1 (5 m)( 3 / 2) (75 kg)(9 . 8 m / s 2 )(3 m) 1 2 + (10 kg)(9 . 8 m / s 2 )(2 . 5 m) 1 2 = 282 N 1 (b) The sum of horizontal forces F x = F w + F g,x = 0. So F g,x =- F w =- 282 N. The sum of vertical forces F y = W c + W l + F g,y = 0. Then F g,y =- (75 kg)(9 . 8 m / s 2 )- (10 kg)(9 . 8 m / s 2 ) = 833 N | F g | = ( F 2 g,x + F 2 g,y ) 1 2 = 879 N (c) The angle of the force is = tan- 1 ( F g,y /F g,x ) = 71 with respect to the horizontal....
View Full Document
This note was uploaded on 01/22/2009 for the course PHYS 207 taught by Professor Liepe during the Fall '08 term at Cornell.
- Fall '08