When objects rest on an inclined plane that makes an angle θ with the horizontal surface, the weight of the object can be resolved into components that act perpendicular ( w⊥ ) and parallel ( w ∥ ) to the surface of the plane.These components can be calculated using:Group of answer choicesw⊥ = w sin (θ) = mg sin (θ) andw∥ = w cos (θ) = mg cos (θ)w ∥ = w sin (θ) = mg sin (θ) andw⊥ = w cos (θ) = mg cos (θ)w ∥ = w cos (θ) = mg sin (θ) andw⊥ = w sin (θ) = mg cos (θ)w⊥ = w cos (θ) = mg sin (θ) andw∥ = w sin (θ) = mg cos (θ)
Question
When objects rest on an inclined plane that makes an angle θ with the horizontal surface, the weight of the object can be resolved into components that act perpendicular ( w⊥ ) and parallel ( w ∥ ) to the surface of the plane.These components can be calculated using:Group of answer choicesw⊥ = w sin (θ) = mg sin (θ) andw∥ = w cos (θ) = mg cos (θ)w ∥ = w sin (θ) = mg sin (θ) andw⊥ = w cos (θ) = mg cos (θ)w ∥ = w cos (θ) = mg sin (θ) andw⊥ = w sin (θ) = mg cos (θ)w⊥ = w cos (θ) = mg sin (θ) andw∥ = w sin (θ) = mg cos (θ)
Solution
The correct answer is:
w⊥ = w cos (θ) = mg cos (θ) and w∥ = w sin (θ) = mg sin (θ)
This is because the component of the weight perpendicular to the inclined plane is given by w⊥ = w cos (θ) = mg cos (θ), and the component of the weight parallel to the inclined plane is given by w∥ = w sin (θ) = mg sin (θ). Here, w is the weight of the object, m is its mass, g is the acceleration due to gravity, and θ is the angle of the inclined plane with the horizontal.
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