A pair of fuzzy dice is hanging by a string from your rearview mirror while you are accelerating from a stoplight to 28m/s in 6.0swhat angle θ does the string make with the vertical?.
2 answers
Yea, I have the same problem except the numbers are a little different does anyone know an equation that could help?
Horizontal components of the forces are as follows:
T(sin Θ) = ma --- Equation 1
and the vertical components of the forces are
T(cos Θ) = mg --- Equation 2
where
T = tension in the string
Θ = angle that string makes with the vertical
m = mass of the object
a = acceleration = 28/6 = 4.667 m/sec^2
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
Dividing Equation 1 with Equation 2,
T(sin Θ)/T(cos Θ) = ma/mg
and simplifying the above,
tan Θ = 4.667/9.8
tan Θ = 0.476
Θ = arc tan 0.476
Θ = 25.5 degrees
T(sin Θ) = ma --- Equation 1
and the vertical components of the forces are
T(cos Θ) = mg --- Equation 2
where
T = tension in the string
Θ = angle that string makes with the vertical
m = mass of the object
a = acceleration = 28/6 = 4.667 m/sec^2
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
Dividing Equation 1 with Equation 2,
T(sin Θ)/T(cos Θ) = ma/mg
and simplifying the above,
tan Θ = 4.667/9.8
tan Θ = 0.476
Θ = arc tan 0.476
Θ = 25.5 degrees
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