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Torque: Overhanging Books on Table

  • Thread starter mochigirl
  • Start date
6
0
1. Homework Statement

3 identical books, each with length L, are stacked over the edge of a table. The top book overhangs the middle book by 1/2L and the middle book overhangs the bottom book by 1/4L. How much can the bottom book be allowed to overhang the edge of the table without the books the (other) books falling?



2. Homework Equations



3. The Attempt at a Solution

I approached this problem assuming that in order for the three books to not fall, their center of gravity must be at least over the table so that there is normal force to counterbalance the books' gravitational force. And so the net torque of the book mass must 0. I placed the pivot point for the torque at the left end of the bottom book

So far, I have \tau=-[(mL/2)+m(L-xL)+m(L+L/4)]/3m * 3mg

x= the fraction of the bottom book's length allowed to overhang the table.
but I have no idea what to do with the normal force and what the normal force value is.

-------ALSO-------

I read from another thread in which people said that as long as the center of mass of the 2nd book is above the edge of the bottom book and the same for the bottom book over the table, the problem is solved. (https://www.physicsforums.com/archive/index.php/t-73791.html)

I tried that on mastering physics with answer being 1/4L and tried again with 1/2L. neither worked...can anyone help me?

Thanks!
 

LowlyPion

Homework Helper
3,079
4
1. Homework Statement

3 identical books, each with length L, are stacked over the edge of a table. The top book overhangs the middle book by 1/2L and the middle book overhangs the bottom book by 1/4L. How much can the bottom book be allowed to overhang the edge of the table without the books the (other) books falling?

I approached this problem assuming that in order for the three books to not fall, their center of gravity must be at least over the table so that there is normal force to counterbalance the books' gravitational force. And so the net torque of the book mass must 0. I placed the pivot point for the torque at the left end of the bottom book

So far, I have \tau=-[(mL/2)+m(L-xL)+m(L+L/4)]/3m * 3mg

x= the fraction of the bottom book's length allowed to overhang the table.
but I have no idea what to do with the normal force and what the normal force value is.
Welcome to PF.

That is the right approach. Toss m*g and treat them all as having unitary weight just for simplicity.

For T = 0 then

(L/2 - x) bottom book
(L/4 - x) middle book
-(L/4 + x) top book (because it's necessarily to the other side of the pivot point.)

Just add those puppies up and know they should = 0. Solve for x in terms of L.
 
6
0
Welcome to PF.

That is the right approach. Toss m*g and treat them all as having unitary weight just for simplicity.

For T = 0 then

(L/2 - x) bottom book
(L/4 - x) middle book
-(L/4 + x) top book (because it's necessarily to the other side of the pivot point.)

Just add those puppies up and know they should = 0. Solve for x in terms of L.
Thanks for replying! I have some questions on how you arrived at the answer.

1) how did you get those center of gravity positions for each book
2) what did you mean by "to other side of the pivot point"
and how do you treat the three books as one unitary mass if we only have the center of gravity points for each book?
 

LowlyPion

Homework Helper
3,079
4
Thanks for replying! I have some questions on how you arrived at the answer.

1) how did you get those center of gravity positions for each book
2) what did you mean by "to other side of the pivot point"
and how do you treat the three books as one unitary mass if we only have the center of gravity points for each book?
1) Well I thought I was reading the drawing that I made from the problem description.

Bottom book only overlaps the pivot by x, what you want to find. It's CoM is at L/2.
Middle book is offset L/4 closer. So the CoM is L/4 closer.
Top book is offset L/2 farther still, which puts its CoM beyond the edge of the table by L/4 and the offset x still to be added to that.

2) I like to simplify so I can focus. All the m*g's acting at the CoM of each book (at L/2 of each book) just want to be gotten rid of to my eye, so that's what I did. Poof. Unit weight acting at L/2.

As to the pivot point, I chose the edge of the table. Things to the left + and things to the right -.
 
6
0
ooh, I see my problem. I set the pivot to the left END of the bottom bottom so I got really confused. I changed the pivot point and used a slightly different method setting the center of mass of the bottom book to x1 and got -L/3

and then did L/2-L/3=L/6

That's the same answer as yours. YAY =D THANK YOU!
 

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