Max Distance: Find Center of Box from Table End

  • Thread starter Thread starter Dark Visitor
  • Start date Start date
  • Tags Tags
    Maximum
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
4 replies · 22K views
Dark Visitor
Messages
215
Reaction score
0
I am unsure of how to go about this problem. So if anyone could help me by going step by step through it, showing all equations and numbers used, and then showing me the answers so I can make sure I get the same thing, I would appreciate it.


A thin 2.50 kg box rests on a 7.00 kg board that hangs over the end of a table, as shown in the figure. How far can the center of the box be from the end of the table before the board begins to tilt?

http://session.masteringphysics.com/problemAsset/1013745/6/jfk.Figure.P08.13.jpg
 
Physics news on Phys.org

Discussion

Dark Visitor said:
I am unsure of how to go about this problem. So if anyone could help me by going step by step through it, showing all equations and numbers used, and then showing me the answers so I can make sure I get the same thing, I would appreciate it.


A thin 2.50 kg box rests on a 7.00 kg board that hangs over the end of a table, as shown in the figure. How far can the center of the box be from the end of the table before the board begins to tilt?

http://session.masteringphysics.com/problemAsset/1013745/6/jfk.Figure.P08.13.jpg
Think of this as two masses the block and the centre of mass of the board applying torque about the fulcrum. What is the condition for tilting? (hint: it has to do with torque about the fulcrum).

AM
 
I am not sure what you mean by fulcrum. I don't think I have ever heard that word before. Sorry. Can you explain it please?
 
What he means by the fulcrum is the point of balance- Imagine a see-saw. The centre is the fulcrum.
Here the fulcrum is the edge of the table.
Since the block is 30 cm on the table and 20 cm outside it, you can take the downward force acting on the left hand side as 3/5 x 7 = 4.2 kg. This acts at the CG which is 15 cm away.
Hence moments on the left side = 4.2 x .15 = 0.63 kg-m.
Similarly moment on the right side = (2/5 x 7) x .1 = 0.28.
To this is added the moment due to the box with a weight of 2.5 kg at a dist of let's say S metres.
So moment on rt side = 0.28 + 2.5 x S.
when it is just about to fall,
0.63 = 0.28 + 2.5 x S.
Find out the value of S.
 
I got it to be .14 m, which turned out to be the right answer. Thanks a lot. You're explanation helped me understand what I was missing. I guess I was going about it all wrong.