Inclined Plane Problem :D (1 Viewer)

[seventhroot]

From Fundamentals of Physics 10th Ed

When I resolve the forces I get this:

but the answers have this:

I can't see how the forces in the y direction is meant to be F_N - mgcosθ = 0

I think the error stems the mg component but I can't see where I am going wrong. inb4can'tdoyear9trigonometry

I know the mgsinθ part is a mistake; it should be mgtanθ

The answers are not incorrect (apparently)

will rep

 

[glittergal96]

From Fundamentals of Physics 10th Ed

When I resolve the forces I get this:

but the answers have this:

I can't see how the forces in the y direction is meant to be F_N - mgcosθ = 0

I think the error stems the mg component but I can't see where I am going wrong. inb4can'tdoyear9trigonometry

I know the mgsinθ part is a mistake; it should be mgtanθ

The answers are not incorrect (apparently)

will rep

Why do you think the part is a mistake? It looks right to me.

Read the paragraph carefully, we are rotating coordinate axes so x is "up the slope" and y is the upwards normal to the slope.

The weight-force contributes in THIS x-direction and in THIS y-direction.

(When resolving a 2-d force vector into two perpendicular components using a triangle, remember that the original force needs to be the hypotenuse of this triangle!)

 

[seventhroot]

Why do you think the part is a mistake? It looks right to me.

Read the paragraph carefully, we are rotating coordinate axes so x is "up the slope" and y is the upwards normal to the slope.

The weight-force contributes in THIS x-direction and in THIS y-direction.

what I did was rotate the problem so that the the Fn axis was the "y axis" and Tension was the "x axis" so that I don't have to worry about splitting the other 2

I thought it was wrong because

(When resolving a 2-d force vector into two perpendicular components using a triangle, remember that the original force needs to be the hypotenuse of this triangle!)

why is this? I think this is where I am going wrong

inb4n00b question; how did you break it up into components?

 

[glittergal96]

I don't get your logic, anyway yeah what the answers do is right.

why is this? I think this is where I am going wrong

inb4n00b question; how did you break it up into components?

Because we are resolving into perpendicular components, so the other two sides need to be perpendicular. Also, it makes sense that each component should be less that the total force in magnitude.

To break it up into components I drew a triangle with hypotenuse the weight force and other two sides parallel and perpendicular to the incline and used basic trig.

 

[seventhroot]

I don't get your logic, anyway yeah what the answers do is right.

Because we are resolving into perpendicular components, so the other two sides need to be perpendicular. Also, it makes sense that each component should be less that the total force in magnitude.

To break it up into components I drew a triangle with hypotenuse the weight force and other two sides parallel and perpendicular to the incline and used basic trig.

so what I did was make the x and y axis in the T and F directions respectively and with my "basic trig" was this:

and it looks right? but that's still not what they have

the hypotenuse is y so would the magnitude of that be greater than mg?

 

[aDimitri]

so what I did was make the x and y axis in the T and F directions respectively and with my "basic trig" was this:

and it looks right? but that's still not what they have

the hypotenuse is y so would the magnitude of that be greater than mg?

how can the magnitude of Y be greater than mg when it is a directional component of mg?
mg is the hypotenuse, the line along which you drew x should be perpendicular to T if you are going to resolve along those directions

 

[glittergal96]

so what I did was make the x and y axis in the T and F directions respectively and with my "basic trig" was this:

and it looks right? but that's still not what they have

the hypotenuse is y so would the magnitude of that be greater than mg?

reread what I said about the two non weight-force sides of the triangle.

also, see what adimitri says above.

 

[SquareHeartsAdrita]

I see where I am going wrong now

I was drawing the triangle wrong lel

thanks to you all; I get what you were all saying now. I "liked" and repped you both