Splitting reaction into horizontal and vertical components

[green-beans]

Homework Statement

In each of the diagrams (please see attached file (I am sorry for the rotated, the original was in normal form but when uploading it, it was somehow rotated)) and the description of each case below, a particle is moving on a smooth surface, so that the reaction force R acting on the particle from the surface is perpendicular to the surface. In each case give the horizontal (i) and vertical (k) components of the reaction force in terms of its unknown magnitude R
(a) Particle on a horizontal table
(b) Particle on a slope inclined at an angle theta to the horizontal
(c) Particle on a slope inclined at an angle theta to the vertical
(d) Particle on the outside of a sphere with latitude described by theta
(e) Particle on the inside of a cylinder, with height above the bottom described by $theta$
(f) Particle on the inside of a paraboloid bowl (bowl surface is given by z = x² +y²), instantaneously at a position where y=0.

Homework Equations

The Attempt at a Solution

(a) R = Rk
(b) R = Rcos(theta)k - Rsin(theta)i
(c) R = Rsin(theta)k - Rcos(theta)i
(d) Here I am not sure how to express R in terms of components since I am not sure how to find the angle formed by the vertical component and R or the angle formed by the horizontal component and R
(e) R = Rcos(theta)k - Rsin(theta)i In this case I feel that I am oversimplifying the situation as the answer looks exactly as in (b)
(f) Here I am not given any angles and so I am not sure how to use the given equation of the surface except for the fact that y=0 yields in z=x².

I would appreciate any help!
Thank you in advance!

 

[TSny]

For (d), make sure you draw $\vec{R}$ perpendicular to the surface.

(e) Looks correct. To relate it to (b), draw a tangent line to the curve of (e) at the location of the particle.

(f) Think about the geometrical interpretation of the derivative of a function.

 

[green-beans]

For (d), make sure you draw $\vec{R}$ perpendicular to the surface.

(e) Looks correct. To relate it to (b), draw a tangent line to the curve of (e) at the location of the particle.

(f) Think about the geometrical interpretation of the derivative of a function.

Thank you for your reply!
For (d) it seems my diagram was not really representative (R wasn't entirely perpendicular) and so now I got that R = Rsin(theta)i + Rcos(theta)k
For (f) I guess R will be in the direction of the normal at the point where the particle is. So, dz/dx = 2x and so the gradient of the normal should be -1/(2x). So the equation of the normal would be z-z0 = (-1/2x) (x-x0). Now the only problem is that I am not given the position of the particle, i.e. I do not know what x0 and z0 are. This is as far as I got unfortunately :(

 

[TSny]

For (d) it seems my diagram was not really representative (R wasn't entirely perpendicular) and so now I got that R = Rsin(theta)i + Rcos(theta)k

OK, for θ measured from the vertical. I'm not sure how you are supposed to interpret "latitude" on the sphere. Is it the angle measured from the vertical or from the horizontal?

For (f) I guess R will be in the direction of the normal at the point where the particle is. So, dz/dx = 2x and so the gradient of the normal should be -1/(2x). So the equation of the normal would be z-z0 = (-1/2x) (x-x0). Now the only problem is that I am not given the position of the particle, i.e. I do not know what x0 and z0 are. This is as far as I got unfortunately :(

$\vec{R}$ will depend on the position of the particle. So, you will need to express $\vec{R}$ in terms of $x$, say, where $x$ is the $x$-coordinate of the particle. If you let θ be the angle between $\vec{R}$ and the vertical, can you express the trig functions of θ in terms of $x$?

 

[green-beans]

OK, for θ measured from the vertical. I'm not sure how you are supposed to interpret "latitude" on the sphere. Is it the angle measured from the vertical or from the horizontal?

$\vec{R}$ will depend on the position of the particle. So, you will need to express $\vec{R}$ in terms of $x$, say, where $x$ is the $x$-coordinate of the particle. If you let θ be the angle between $\vec{R}$ and the vertical, can you express the trig functions of θ in terms of $x$?

For (d) the angle measured by θ is the angle between the vertical and the line joining the centre of the semi circle with the centre of the particle (as it is indicated on the picture) :)
For (f) I am still a little bit confused as I am getting something really weird. If I let (a,b) be the position of the particle and θ the angle between the vertical and R, then I will have sinθ = R/a (which is horizontal component). As for the vertical component I am not sure how to find it since we do not know the point where R meets the vertical and so I decided to draw a symmetrical triangle on the left hand side from the vertical so that the angle now becomes 2θ. Then I apply the rule of cosines, that is:
(2a)²=R²+R² - 2Rcos2θ
By simplifying this I obtain
cos²θ = (Rsin²θ - 2R)/(2sin²θ) + 1/2
Then I plug in the expression for sinθ = R/a to get that
cos²θ = (R²-2a²+Ra)/(2Ra)
And so to find cosθ, I then should take a square root. But from what I seem to get I think I am doing something wrong.

 

[TSny]

For (f) I am still a little bit confused as I am getting something really weird. If I let (a,b) be the position of the particle and θ the angle between the vertical and R, then I will have sinθ = R/a (which is horizontal component).

I don't understand this. $R$ is a force while $a$ is a position coordinate. The ratio $R/a$ doesn't have the proper units to represent sinθ (which should be dimensionless).

The derivative of the function is related to the angle of tilt of the tangent line to the curve. So, you should be able to determine sinθ and cosθ by using the result for the derivative.

 

[green-beans]

I don't understand this. $R$ is a force while $a$ is a position coordinate. The ratio $R/a$ doesn't have the proper units to represent sinθ (which should be dimensionless).

The derivative of the function is related to the angle of tilt of the tangent line to the curve. So, you should be able to determine sinθ and cosθ by using the result for the derivative.

Ohhh, okay, I got the answer! Thank you!