Re: Question about falling bodies

From: John C. Polasek (jpolasek_at_cfl.rr.com)
Date: 02/12/05 

Date: Sat, 12 Feb 2005 15:33:15 -0500

On Sat, 12 Feb 2005 12:58:01 +1100, "Rob Bruce"
<robbruce@ozemail.com.au> wrote:

>Hi all,
>
>I need some help answering a pretty basic question that I just can't seem to
>get right.
>
>I have a mass of 100kg attached to a rope. The rope has been shown to break
>if a force of 4200N is applied to it.
>
>How far does the mass have to fall to exert a force of 4200N on the rope (I
>don't know the length of the rope but am allowed to assume the 4200N force
>will be able to build up before the rope applies a breaking force - i.e. I
>can assume that the rope is just long enough for the minimum force needed to
>snap it to be built up by the falling mass).
>
>Mass starts at rest (i.e. initial velocity =0)
>
>I am allowed to ignore all forces except gravity (i.e no angular forces mass
>is assumed to fall directly to earth, no air resistance etc).
>
>I started off thinking momentum was the key and went through the classical
>formulas to back calculate the velocity and displacement but that strikes me
>as too simplistic.
>
>So I throw myself on the mercy of the group to give me some thoughts.....
>
>thanks
>
>Rob
>
Yes you can solve this if you first hang the 100kg or 980 Newtons
weight W on the rope and find deflection D1 say .01m.
        D1 = .01m = .4"
The rope has elastic coefficient k:
        W = k*D1= 980N k = W/D1= 9.8x10^4N/m
The ratio of breaking force to actual weight is
        n = 4200/980 = 4.286
        D2 = nD1
Drop from height H and energy
        WH = .5kD2^2.
 
 and find finally that H = .5*D1*n^2, or .092m = 3.6".
        H/D1 = n^2/2
The drop height is almost nothing, being the original test D1 times
n^2/2, with n the ratio of breakweight to actual weight.

Mr. Dual Space

If you have something to say, write an equation.
If you have nothing to say, write an essay

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