Asked by Ashley
A large kite of mass 3 kg is flying through the air on a windy day. Currently, the tension from the string on the kite has a magnitude of 7.7 N at an angle of θ = 33.5 degrees. The current acceleration of the kite has a magnitude of a = 7.56 m/s2 at an angle of φ = 38.5 degrees. The only forces felt by the kite are its own weight, the tension from the string, and a force from the wind. Find the X and Y component of the force of the wind.
Find the X component and Y component of the force of the wind on the kite
Find the X component and Y component of the force of the wind on the kite
Answers
Answered by
Elena
There are three forces acting on the kite:
1. gravity
m•g(x) = 0
m•g(y) = m•g=3•9.8 =29.4 N.
2. tension
T(x) = T•sinθ=7.7•sin33.5° = 4.25 N,
T(y) = T•cosθ=7.7•cos 33.5° =6.42 N.
3. wind force
m•a(x) = - m•a•cosφ = - 3•7.56•cos38.5°= - 17.75 N
m•a(y) = m•a•sinφ =3•7.56•sin38.5°= 14.12N
Total force F:
F(x) = m•g(x)+ T(x)+ m•a(x) =0+4.25-17.75 = - 13.5 N,
F(y) = m•g(y)+ T(y)+ m•a(y) =29.4+6.42+14.12 = 49.94 N.
1. gravity
m•g(x) = 0
m•g(y) = m•g=3•9.8 =29.4 N.
2. tension
T(x) = T•sinθ=7.7•sin33.5° = 4.25 N,
T(y) = T•cosθ=7.7•cos 33.5° =6.42 N.
3. wind force
m•a(x) = - m•a•cosφ = - 3•7.56•cos38.5°= - 17.75 N
m•a(y) = m•a•sinφ =3•7.56•sin38.5°= 14.12N
Total force F:
F(x) = m•g(x)+ T(x)+ m•a(x) =0+4.25-17.75 = - 13.5 N,
F(y) = m•g(y)+ T(y)+ m•a(y) =29.4+6.42+14.12 = 49.94 N.
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