Question

assuming no sliding and that the shoulder is 1.2m from the feet,what force is reqiuered to topple a 70 kg person standing with his feet spread 0.9 m? and can i have an explination for the answer?..thanks

Answer 1

assuming no sliding and that the shoulder is $1.2\mathrm{m}$ from the feet, what force is required to topple a $70\mathrm{kg}$ person standing with his feet spread $0.9\mathrm{m}$ ? and can i have an explanation for the answer?

Solution

Make sketch:

Assuming that the toppling force is horizontal ant applied from the side.

Weight of the person is:

$$W = m * g = 7 0 k g * 9. 8 1 N / k g = 6 8 6. 7 N$$

For point A: The torque caused by weight is:

$$\tau_ {W} = W * l _ {W}$$

Where $l_{W}$ is :

$$l _ {W} = 0. 9 / 2 = 0. 4 5 m$$

The torque caused by your force is:

$$\tau_ {F} = F * l _ {F}$$

Where

$$l _ {F} = 1.2 \, \text{m}$$

So force required to topple person is:

$$\tau _ {F} = \tau _ {W}$$ $$F * l _ {F} = W * l _ {W}$$ $$F = W * l _ {W} / l _ {F}$$

Calculating:

$$F = 686.7 * 0.45 / 1.2 = 257.5 \, \text{N}$$

Answer: 257.5 N