{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW_5_solutions

HW_5_solutions - P112 S07 Solutions to HW#5 1 2 kg 90 A = m...

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
This is the end of the preview. Sign up to access the rest of the 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) Free-body diagram for the barbell: Free-body 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 third-law 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

{[ snackBarMessage ]}

Page1 / 4

HW_5_solutions - P112 S07 Solutions to HW#5 1 2 kg 90 A = m...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online