What Angles Affect the Vector Components of a Weight on an Inclined Plane?

[dandy9]

Homework Statement

The weight of a book sliding down a frictionless inclined plane can be broken into two vector components: one acting parallel to the plane, and the other acting perpendicular to the plane.
(a) At what angle is the components equal?
(b) At what angle is the component parallel to the plane equal to zero?
(c) At what angle is the component parallel to the plane equal to the weight?

Homework Equations

Fnet = ma

The Attempt at a Solution

I'm not really sure how to approach this.
My guess for (a) is 45 just because it sounds reasonable that the components would be equal at the middle of 0 to 90 degrees.

 

[cupid.callin]

try to imagine a book on a inclined plane inclined at say angle a...
If weight is mg, component of weight perpendicular to incline is ..?
what is the component of weight parallal to incline?

draw figure and then you'll easily solve it

 

[dandy9]

Thank you!

I got them all - thanks for getting me started! This is how I did it:
(a) cos and sin are equal at 45degrees
(b) the sin of 0 gives a component of 0.
(c) if the component and weight are equal then you take the inverse sin of 1 and get 90.

 

[cupid.callin]

Very good.

 

[rwisz]

For (b) I'm not sure, maybe 90 degrees? And for (c) I'm also not sure, maybe 0 degrees?

As a scientist, it is important to approach problems with a clear understanding of the relevant equations and principles. In this case, the relevant equation is Newton's second law, Fnet = ma, which tells us that the net force acting on an object is equal to its mass multiplied by its acceleration. This can be applied to the book sliding down the inclined plane.

(a) To determine the angle at which the components are equal, we can use trigonometry to break down the weight (W) of the book into its components. The component parallel to the plane (Fpar) can be found using the equation Fpar = Wsinθ, where θ is the angle of the incline. The component perpendicular to the plane (Fperp) can be found using the equation Fperp = Wcosθ. Setting these two components equal to each other, we can solve for θ: Wsinθ = Wcosθ. Simplifying, we get tanθ = 1, or θ = 45 degrees. Therefore, the components are equal at an angle of 45 degrees.

(b) To find the angle at which the component parallel to the plane is equal to zero, we can set Fpar = 0 and solve for θ: 0 = Wsinθ. This equation has infinite solutions, as any angle whose sine is equal to 0 will satisfy it. Therefore, the component parallel to the plane is equal to zero at any angle where the incline is completely vertical, or 90 degrees.

(c) Finally, to find the angle at which the component parallel to the plane is equal to the weight, we can set Fpar = W and solve for θ: W = Wsinθ. Simplifying, we get sinθ = 1, or θ = 90 degrees. Therefore, the component parallel to the plane is equal to the weight at an angle of 90 degrees, or when the incline is completely horizontal.