Assuming no sliding and that the shoulder is 1.2m from the feet, what force is required to topple a person weighing 70 Kg standing with his feet spread 0.9 m? If possible, please include an explanation about your answer.

  • $\begingroup$ I'd imagine there would be a difference between what steady force is needed and what would be needed as a sudden impulse catching them off guard. Also, are we toppling them over backwards or sideways? $\endgroup$ – Michael Brown Jan 11 '13 at 11:25
  • $\begingroup$ i don't really know that is what the problem says and nothing more..but let's say sideway if it is needed $\endgroup$ – roza Jan 11 '13 at 11:29
  • $\begingroup$ Is this a homework problem? $\endgroup$ – Michael Brown Jan 11 '13 at 11:30
  • $\begingroup$ not really..i was reading a physic book of a friend and this problem took my interest but i just can't get the answer and it is driving me crazy $\endgroup$ – roza Jan 11 '13 at 11:32
  • 1
    $\begingroup$ rules of static equilibrium,i believe it should be those rules $\endgroup$ – roza Jan 11 '13 at 11:36

Approximate the person with a brick with a width of 0.9 m and a height of 1.2 meter. The torque around the tipping point caused by gravity is $mg\cdot l$, where $l$ is the horizontal distance from the tipping point to the center of mass of the brick, i.e. half the brick width assuming it has a uniform density. You need to counteract this torque by exerting a horizontal force at the top. The torque caused by your force is $F\cdot h$. To sum up:

$$ mgl=Fh \Rightarrow F=\frac{mgl}{h}\approx\frac{70\cdot9.8\cdot0.45}{1.2}\approx257 \mathrm{N} $$

However, many of the assumptions made here may not be realistic in a real world person-toppling event.

Edit: The solution above assumes that the toppling force is horizontal. If you can apply the force at an angle you can get a slightly longer lever, maximally the length of the brick diagonal: $$ d=\sqrt{1.2^2+0.9^2}=1.5 $$ This gives the smallest possible toppling force: $$ F=\frac{mgl}{d}\approx\frac{70\cdot9.8\cdot0.45}{1.5}\approx206 \mathrm{N} $$

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