# ch4 - F m y ∑ =(2 88 2 9 N kg − = T a f F G g = 88 2 N...

This preview shows pages 1–3. Sign up to view the full content.

4.17 4.19 4.22

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4.28 First, consider the block moving along the horizontal. The only force in the direction of movement is T . Thus, Fm x = a a a a (1) T = 5 kg a f Next consider the block that moves vertically. The forces on it are the tension T and its weight, 88.2 N. We have Fm
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: F m y ∑ = (2) 88 2 9 . N kg − = T a f F G g = 88 2 . N T G T G n G FIG. P4.28 Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be added to give 88 2 14 . N kg = b a g . Then 4.43 4.45 4.54...
View Full Document

## This note was uploaded on 02/15/2011 for the course PHYS 6A taught by Professor Mahaashour-abdalla during the Summer '07 term at UCLA.

### Page1 / 3

ch4 - F m y ∑ =(2 88 2 9 N kg − = T a f F G g = 88 2 N...

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

View Full Document
Ask a homework question - tutors are online