Answer:
2.52 m/s
Explanation:
When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.
Sum of the forces in the radial direction:
∑F = ma
mg = m v² / r
g = v² / r
v = √(gr)
Given that r = 0.650 m:
v = √(9.8 m/s² × 0.650 m)
v = 2.52 m/s
The slope of the line is
(change in ' y ' between the ends) / (change in ' x ' between the ends)
Slope = (630g - 0) / (70 cm^3 - 0)
Slope = (630 / 70) g/cm^3
<em>Slope = 9.0 g/cm^3</em>
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
the third law
Explanation:
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