{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw29_s - homework 29 YOO HEE Due 4:00 am Question 1 chap 14...

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

homework 29 – YOO, HEE – Due: Apr 10 2008, 4:00 am 1 Question 1, chap 14, sect 1. part 1 of 4 10 points A uniform brick of length 20 cm is placed over the edge of a horizontal surface with the maximum overhang x possible without falling. g x 20 cm Find x for a single block. Correct answer: 10 cm (tolerance ± 1 %). Explanation: Let : L = 20 cm For n bricks the center of mass is x m n summationdisplay i =1 x i m i n summationdisplay i =1 m i = 1 n n summationdisplay i =1 x i , where x i is the center of mass position of the i th brick and m i is the mass of the i th brick. Since x 1 L = 1 2 , as measured from the maxi- mum overhang, the center of mass of a single brick is in its middle or x m L vextendsingle vextendsingle vextendsingle n =1 = 1 2 1 = 1 2 of the brick’s length from its maximum over- hang. x m = 1 2 L = 1 2 (20 cm) = 10 cm . g x 20 cm Question 2, chap 14, sect 1. part 2 of 4 10 points Two identical uniform bricks of length 20 cm are stacked over the edge of a hori- zontal surface with the maximum overhang x possible without falling. g x 20 cm Find x for two blocks. Correct answer: 15 cm (tolerance ± 1 %). Explanation: Since m i = m (all bricks have the same mass), n summationdisplay i =1 m = n m . The bricks will just balance when the center of mass is over the fulcrum; i.e. , the edge of the horizontal sur- face. Measurement will be made from the left edge of the top brick with the maximum overhang. To calculate the center of mass x m L = 1 n L n summationdisplay i =1 x i ; when an additional brick is positioned under the stack of n - 1 bricks, the additional brick’s left edge is placed at the balance point of the previous stack of n bricks. The center of mass of the additional brick is 1 2 of a brick’s length plus the maximum overhang of the previous stack of n - 1 bricks. The top brick can extend 1 2 of a brick’s length from the maximum overhang. Since x 2 L = x m L vextendsingle vextendsingle vextendsingle n =1 + 1 2 = 1 2 + 1 2 = 1 as measured from the maximum overhang, the center of mass of the two bricks is in their middle or x m L vextendsingle vextendsingle vextendsingle n =2 = 1 2 + 1 2 = 3 4 of a single brick’s length from the maximum overhang: x m = 3 4 L

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

View Full Document
homework 29 – YOO, HEE – Due: Apr 10 2008, 4:00 am 2 = 3 4 (20 cm) = 15 cm . g x 20 cm Question 3, chap 14, sect 1. part 3 of 4 10 points Three identical uniform bricks of length 20 cm are stacked over the edge of a hori- zontal surface with the maximum overhang x possible without falling. g x 20 cm Find x for three blocks. Correct answer: 18 . 3333 cm (tolerance ± 1 %).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

hw29_s - homework 29 YOO HEE Due 4:00 am Question 1 chap 14...

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

View Full Document
Ask a homework question - tutors are online