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An 85 kg man stands in a very strong wind moving at 15 m/s at torso height. As you know, he will need to lean in to the wind, and we can model the situation to see why. Assume that the man has a mass of 85 kg, with a center of gravity 1.0 m above the ground. The action of the wind on his torso, which we approximate as a cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground, produces a force that tries to tip him over backward. To keep from falling over, he must lean forward.

Required:
a. What is the magnitude of the torque provided by the wind force? Take the pivot point at his feet. Assume that he is standing vertically.
b. At what angle to the vertical must the man lean to provide a gravitational torque that is equal to this torque due to the wind force?

Sagot :

okpalawalter8

Answer:

  a) [tex]T= 72.9 Nm[/tex]

  b)      [tex]\Theta = 5 \textdegree[/tex]

Explanation:

From the question we are told that

Mass 85kg

Speed 15m/s

Center of gravity 1.0m

Cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground

a)Generally equation for force is given by

   [tex]F = 0.5 p A v^2[/tex]

Mathematically solving for force exacted

   [tex]F = 0.5*1.2*0.9*0.5*15^2[/tex]

   [tex]F= 60.75 N[/tex]

Mathematically solving for torque

Torque,

    [tex]T = r * F[/tex]

    [tex]T=1.2*60.75[/tex]

   [tex]T= 72.9 Nm[/tex]

b)Generally in solving for [tex]\theta[/tex]

   [tex]Tan\theta = Torque/(Mass * Gravity)[/tex]

   [tex]Tan\theta = (72.9)/(85 * 9.8)[/tex]

     [tex]\Theta = 5 \textdegree[/tex]

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