A body of mass 6kg rests on an inclined plane. The normal reaction is R, and the limiting frictional force is F as shown in the image above. if F = 30N, and g = 10ms\(^{-2}\) then the angle of inclination \( \theta \) is
Calculate the weight of the body: \(W = mg = 6 \, \text{kg} \times 10 \, \text{m/s}^2 = 60 \, \text{N}\)
The limiting frictional force balances the component of weight acting down the incline: \(F = W \sin \theta\)
Substituting the known values: \(30 = 60 \sin \theta\)
Solving for \( \sin \theta \): \(\sin \theta = \frac{30}{60} = \frac{1}{2}\)
The angle \( \theta \) corresponding to \( \sin \theta = \frac{1}{2} \) is: \(\theta = 30^\circ\)
The angle of inclination \( \theta \) is \(30^\circ\).
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}