A car weighs 2,000 kg. It moves along a road by applying a force on the road with a parallel component of 560 N. There are two p
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The hoop is attached.
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>