Mass/height/gravity equation

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In summary, the tower falls to the ground because of the mass, height, and gravity acceleration. The only equation that appears to be physically correct is (d), in which case the time it takes for the tower to fall to the ground is proportional to the length of the tower divided by the gravity acceleration.
  • #1
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Homework Statement



The ground around the Leaning Tower of Pisa one day gives way and the tower topples over. It is found from Mechanics theory that the only things that may determine the time it takes the toppling tower to fall to the ground are:

the mass (m) of the tower
the height (h) of the tower
the gravity acceleration (g).

Given that k is a dimensionless constant, the dependence of the time t for the tower to strike the ground upon m, h and g is given by:

(a) [tex]t=k\sqrt{\frac{g}{h}}[/tex]

(b) [tex]t=k\sqrt{\frac{hm}{g}}[/tex]

(c) [tex]t=k\sqrt{\frac{h}{g}}[/tex]

(d) [tex]t=km\sqrt{\frac{h}{g}}[/tex]

(e) [tex]t=k\sqrt{\frac{m}{g}}[/tex]


The Attempt at a Solution



I don't recognize any of the formulas here so I can't tell which one is the correct answer. The only formula I can think of that interrelates some of these variables is [tex]h=v_y-gt[/tex] and I can replace the v by at to get [tex]h=t(a-g)\Rightarrowt=\frac{a-g}{h}[/tex]. But again the resulting formula doesn't take mass into account. Can anyone help?
 
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  • #2
Try analyse the units (time, length, mass) involved in the different equations (hint: only one of the equations have the same effective unit on both sides).
 
  • #3
Filip Larsen said:
Try analyse the units (time, length, mass) involved in the different equations (hint: only one of the equations have the same effective unit on both sides).

Alright, I have

T=kmxhygz

T1=mxLy(L/T2)z

T1=mxLy(L/T2)z

T1=mxLy Lz T-2z

So we have x+y-z=1. But I'm stuck here and I have an exam tomorrow...
Could you show me how to work out the values of x,y,z? I don't get it! :(
 
  • #4
It is simpler than you think - you just have to check in each case if the dimension on the left-hand side of the equation equals the resulting dimension on the right-hand side. (Note, that when I said "unit" in my earlier answer I really meant "dimension", which looks like how you read it anyway).

For instance, in (a) you have dimension T (time) on the left-hand side (all four equations have that dimension on the left-hand side), and on the right-hand side you have g with dimension LS-2 (length divided by time squared) and h with dimension L, so g/h is of dimension LS-2 L-1 = S-2 and finally after the square root you get square root S-2 = S. Since k is dimensionless, the left-hand T is not clearly not equal to S-1 and so you can conclude that this equation cannot be physically correct.

Similar analysis can be done for the remaining three equations to reveal that for one of them the dimensions are equal and this equation then could express a physical correct relationship between the physical quantities involved. Note, that having an equation where the dimensions check is a required, but in general not a sufficient condition for a physical correct relationship. In this problem, however, it is assumed that at least one equation is physical correct and you then use dimensional analysis to show how to identify that one equation.

Good luck with your exam.
 

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