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pishuonlain [190]
3 years ago
5

A 90 kg man stands in a very strong wind moving at 17 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 90 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.
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. Assume that the air is at standard temperature and pressure.
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?
Physics
1 answer:
victus00 [196]3 years ago
5 0

Answer:

a)  t=195.948N.m

b)  \phi=13.6 \textdegree

Explanation:

From the question we are told that:

Density \rho=1.225kg/m^2

Velocity of wind v=14m/s

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient \mu=2.05

a)

Generally the equation for Force is mathematically given by

F=\frac{1}{2}\muA\rhov^2

F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2

F=163.29

Therefore Torque

t=F*r*sin\theta

t=163.29*1.2*sin90

t=195.948N.m

b)

Generally the equation for torque due to weight is mathematically given by

t=d*Mg*sin90

Where

d=sin \phi

Therefore

t=sin \phi*Mg*sin90

195.948=833sin \phi

\phi=sin^{-1}\frac{195.948}{833}

\phi=13.6 \textdegree

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Answer:

(a) T= 38.4 N

(b) m= 26.67 kg

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Kinematics

d= v₀t+ (1/2)*a*t² (Formula 2)

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

v₀=0, d=18 m , t=5 s

We apply the formula 2 to calculate the accelerations of the blocks:

d= v₀t+ (1/2)*a*t²

18= 0+  (1/2)*a*(5)²

a= (2*18) / ( 25) = 1.44 m/s² to the right

We apply Newton's second law to the block A

∑Fx = m*ax

60-T = 15*1.44

60 - 15*1.44 = T

T = 38.4 N

We apply Newton's second law to the block B

∑Fx = m*ax

T = m*ax

38.4 = m*1.44

m= (38.4) / (1.44)

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