Inclined Plane Angle

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MedPR

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Question 1: Why/how do you know that theta between the incline and the horizontal is the same as theta between the mg vector and the mgcostheta vector?

Edit:

Might as well ask this question here too.

Question 2: When asked to find the equation that represents the acceleration of a package sliding down a frictionless inclined plane that makes an angle theta with the horizontal ground, will this method always work?

Fnet = mg + Fnormal.

Since mg = mgcostheta + mgsintheta

Fnet = mgcostheta+mgsintheta+Fnormal

And since Fnormal and mgcostheta cancel each other out,

Fnet = mgsintheta = ma

so a = gsintheta

Question 3: And if a=gsintheta, then the force acting on the package down the incline is F=mgsintheta?

I'm sure this is basic inclined plane derivation, but I haven't been in Physics for a few years, and I'm realizing now that I might not have learned it (even though we were required to).

Thank you!

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Last edited:
Question 1: Why/how do you know that theta between the incline and the horizontal is the same as theta between the mg vector and the mgcostheta vector?

Try this if it works ... its all about angles:

Jwq0g.jpg


Edit:

Might as well ask this question here too.

Question 2: When asked to find the equation that represents the acceleration of a package sliding down a frictionless inclined plane that makes an angle theta with the horizontal ground, will this method always work?

Fnet = mg + Fnormal.

Since mg = mgcostheta + mgsintheta

Fnet = mgcostheta+mgsintheta+Fnormal

And since Fnormal and mgcostheta cancel each other out,

Fnet = mgsintheta = ma

so a = gsintheta

yes and also when no rope Tension) attached to the object

Question 3: And if a=gsintheta, then the force acting on the package down the incline is F=mgsintheta?


Yes
 
alternative way to look at it:
assuming the incline is frictionless

in the x direction:
Fnet= Fapplied- Fresisting= ma
so
Fnet= parallel force- 0= ma
Fnet=mgsintheta= ma

so Fnet= mgsintheta and gsintheta=a
 
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