This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: P112 S07 Solutions to HW #5 1. 2. kg . 90 A = m . N 490 B on E = W . m 60 . B = ! y in a time s 6 . 1 = ! t . (a) Freebody diagram for the barbell: Freebody diagram for the athlete: Forces shown: B on A F r : contact force due to athlete’s hands acting on barbell. B on E W r : gravitational force due to Earth acting on barbell (weight). A on G N r : normal force due to ground acting on athlete’s feet. y + A on E W r : gravitational force due to Earth acting on athlete (weight). A on B F r : contact force due to barbell acting on athlete’s hands. (b) We want to find the force that the athlete’s feet exert on the ground, G on A N r , as he lifts the barbell. This force is the Newton’s thirdlaw interaction partner of A on G N r , so A on G G on A N N r r ! = . Thus, we need to find the magnitude of A on G N r . Newton’s 2 nd law applied to the athlete yields , A A A on B A on E A on G , A on B , A on E , A on G = = ! ! = + + y y y y a m F W N F W N , because the athlete is not accelerating....
View
Full Document
 Spring '07
 LECLAIR,A
 mechanics, Force, Cos, µk sin

Click to edit the document details