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[Solved] A car is moving on a plane inclined at 30° to the horizontal with an acceleration of 10 ms^–2 parallel to the plane upward.

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A car is moving on a plane inclined at 30° to the horizontal with an acceleration of 10 ms–2 parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is______. (Take g = 10 ms–2)

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tan(30 + θ) = \(\frac{mg \ sin 30° + ma}{mg\ cos 30°}\)

tan(30 + θ) = \(\frac{5+10}{5 \sqrt 3}\) = \(\frac{1+2}{\sqrt3}\)

\(\frac{tan \theta + \frac{1}{\sqrt 3}}{1 - \frac{1}{\sqrt3} tan \theta}\) = √3

√3 tanθ + 1 = 3 - √3 tanθ

2√3 tanθ = 2

tanθ = \(\frac{1}{\sqrt 3}\)

θ = 30°

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