Physics Help Forum Mechanics

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Apr 15th 2018, 12:02 PM #1 Junior Member   Join Date: Apr 2018 Posts: 3 Mechanics Four same homogenious bricks are stacked on top of each other as illustrated below. They all have the horisontal sidelength of ”a”. The bricks are not rotated in comparison to each other. What is the maximum value of the distance ”d” that possibly can be achieved? A illustrative picture of the system: https://i.imgur.com/ITYlGhE.jpg I am aware that the question obviously has to do with balancing the forces. The best answer of mine is d=15*a/16. The correct answer is d=11*a/12. I need help with deriving the correct answer.
 Apr 15th 2018, 01:55 PM #2 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 123 Mechanics You must begin from the top down. Putting the maximum overhang at the top is the best way to optimize this scenario. Every time you go down a level you must make sure that the load is balanced, i.e. half the load sits on the left hand side of the fulcrum (corner of the brick) and the other half must sit on the right hand side. Therefor the first overhang distance will be 1/2 of A. The distance or overhang to get the top two bricks balanced on the third brick is 1/4. Finally the distance or overhang to get the top three bricks balanced on the fourth brick is 2/12. Adding up the successive overhangs gets you to 11/12 of A.
 Apr 15th 2018, 02:55 PM #3 Junior Member   Join Date: Apr 2018 Posts: 3 Thanks dor your reply. I got that the second brick should be layed at the firat ones fulcrum, although not the reasoning behind the third one, and therefore, also fourth. What is the precise reasoning that it is 1/4 and not 1/2? I would love to see how you reason, step by step.
 Apr 15th 2018, 03:23 PM #4 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 123 Mechanics W is the weight of one brick. On the top level you need to balance 0.5W followed by 1W on the second level and on the third level 1.5W. It is 1/4 because the height above the 1/4 increment is two bricks so the weight is doubled. On the third level it becomes 1/6 (2/12) because there are three bricks above. '  ------------------------------- '  |******************************| '  |******************************| ' ------------------------------- ' <-0.5W----><----0.5W--> '1/21/2 '------------------------------- '|**************** **************| '|**************** **************| '------------------------------- '<--------------------> <--------------------> 'W <---->W ' 1/4 '------------------------------- '|************************* *****| '|************************* *****| '------------------------------- '<----------------------> <------------------------> '1.5W 1.5W ' <2/12> '------------------------------- '|******************************| '|******************************| '`------------------------------- Last edited by YellowPeril; Apr 15th 2018 at 03:37 PM.
 May 3rd 2018, 10:42 AM #5 Junior Member   Join Date: Apr 2018 Posts: 3 Wonderful illustration but may you show me the physics? I am aware of the formula M=F*L.
 May 3rd 2018, 03:11 PM #6 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 123 Mechanics You have to work from the top down. Do the calculation in three stages. Use center of gravity calculations. You can calculate X and Y COG but actually only the X COG is important. For an object to balance the X COG needs to rest directly on the fulcrum. In this case you can consider the corner of the block below to be the fulcrum. Here is an example of working out the X COG. Block 1 top to block 4 on the bottom. COG Block 1 = a/2+X Distance X is the distance that block 1 is moved from the left edge of block 4. COG Block 2 = a/2 + Y Distance Y is the distance that block 2 is moved from the left edge of block 4 etc. COG Block 3 = a/2 + Z Distance Z is the distance that block 3 is moved from the left edge of block 4 For Block 1 not to tip on block 2 X must be less than or equal to Y + a/2. i.e. X + a/2 <= Y+ a; X < = Y + a/2 We decide to take X at it's maximum i.e. Y+a/2. We do this so that the Y center of gravity is as low as possible. (Not a completely rigorous approach but should be good enough to satisfy the question) Then the center of gravity of Block 1 and 2 must be calculated as a unit. COG of a combined unit is the weighted average of the coordinates and the mass in X and Y coordinates respectively (Here just the X is will do because it is a balance problem) The combined COG of block 1 and 2 will be (a +Y + X) /2 = (a + Y + Y + a/2)/2 = 3*a/4+Y. Here this value must be less than or equal to Z + a in order for block 1 and 2 not to tip on block 3. Maximum Y can be 1/4a+Z COG of block 1,2 and 3 will be (2*(a+z)+z+a/2)/3 = 5/6*a+z this cannot be further than a. Maximum of Z is 1/6*a. Maximum Y = (1/6+1/4)a = 5/12*a Maximum X = Y+a/2 Maximum X = 11/12.
 May 4th 2018, 03:53 PM #7 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 123 Mechanics You can also use back substitution and then you do not need to make any initial assumptions which makes tha approach a bit more rigorous. COG of block 1 is X+a/2 COG of block 1 and 2 is (X + a/2 + Y + a/2)/2 = (X + Y + a)/2 COG of block 1,2 and 3 is (X+Y+Z + 3/2a)/3 etc. Fulcrum on which block 1 rests is Y+a Fulcrum on which block 2 rests is Z+a Fulcrum on which block 3 rests is a So if the respective COG may not move further than the fulcrum point then X+a/2 <= Y+a...........................1 (X+Y+a)/2 <= Z+a.....................2 (X+Y+Z+3/2a)/3 <= a...................3 Using 3 Z <= 3/2a - X - Y Substituting Z into 2 (X+Y+a)/2 <= 3/2a - X - Y + a Y <=4/3*a-x Substituting Y into 1 X+a/2<=4/3*a-x+a x <=11a/12 y <=5a/12 z <=2a/12
 May 5th 2018, 07:42 AM #8 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 123 Mechanics One final comment: I think you are (incorrectly) getting 15/16a because you ignoring the formula for F=M*L and just balancing for mass, you need to balance for the weighted mass i.e. the formula for COG. (Think of a scale with sliding weights). What is interesting about this problem is that if you extrapolate it to many blocks resting on each other, you can actually show that as the stack of blocks grows in height, the top block can move quite far away from it's support base. i.e top block can move 2a or 3a even (if the stack of blocks is high enough). However the combined center of gravity of the pile of blocks must always rest inside of the support base and this must also be true for an individual portion of the stack taken from the any selected block up to top of the stack.

 Tags mechanics

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post charlielife Thermodynamics and Fluid Mechanics 3 Jun 13th 2014 05:37 AM kude Kinematics and Dynamics 3 Mar 17th 2014 04:41 AM Tyson Advanced Mechanics 3 Dec 3rd 2012 03:13 PM tanaki Kinematics and Dynamics 7 Oct 27th 2009 07:02 AM Ja]v[es Advanced Waves and Sound 1 Mar 19th 2009 01:49 AM