# Thread: Exam Tomorrow! Mechanics quick question!

1. ## Exam Tomorrow! Mechanics quick question!

A particle is on a smooth incline x degrees to the horizontal.

I KNOW the acceleration will be $gsinx$ if the slope surface is smooth but I do not understand how this is formed.

Triginometrically I always get a force of something like $\frac{g}{sinx}$. How do I actually get the acceleration?

Thanks guys!

2. Originally Posted by anthmoo
A particle is on a smooth incline x degrees to the horizontal.

I KNOW the acceleration will be $gsinx$ if the slope surface is smooth but I do not understand how this is formed.

Triginometrically I always get a force of something like $\frac{g}{sinx}$. How do I actually get the acceleration?

Thanks guys!
When in doubt, try Newton's 2nd Law.

I am picturing an inclined plane, sloping downward to the right. I am choosing a +x direction down the slope and a +y direction (which isn't actually needed here) perpendicular to and coming out of the plane. I'm going to call the angle of incline $\theta$ measured between the horizontal and the incline of the plane.

Stick an object of mass m on the plane and assume no friction. The object will begin to slide down the plane. Do a Free-Body Diagram. You'll have a normal force (N) in the +y direction, and a weight (w) straight down.

We need to break w into components in the x and y directions. (Specifically all we care about here is the component in the x direction.)
$w_x = w \cdot sin( \theta )$ <-- Make sure you understand why this is sine and not cosine!

Since w = mg: $w_x = mg \cdot sin( \theta )$.

Newton's 2nd in the x - direction:
$\sum F_x = w_x = ma$

$mg \cdot sin( \theta ) = ma$

$a = g \cdot sin( \theta )$

-Dan

3. Originally Posted by topsquark
When in doubt, try Newton's 2nd Law.

I am picturing an inclined plane, sloping downward to the right. I am choosing a +x direction down the slope and a +y direction (which isn't actually needed here) perpendicular to and coming out of the plane. I'm going to call the angle of incline $\theta$ measured between the horizontal and the incline of the plane.

Stick an object of mass m on the plane and assume no friction. The object will begin to slide down the plane. Do a Free-Body Diagram. You'll have a normal force (N) in the +y direction, and a weight (w) straight down.

We need to break w into components in the x and y directions. (Specifically all we care about here is the component in the x direction.)
$w_x = w \cdot sin( \theta )$ <-- Make sure you understand why this is sine and not cosine!

Since w = mg: $w_x = mg \cdot sin( \theta )$.

Newton's 2nd in the x - direction:
$\sum F_x = w_x = ma$

$mg \cdot sin( \theta ) = ma$

$a = g \cdot sin( \theta )$

-Dan

Ohhh...hmm are you saying here that the accelleration is calculated in the x direction and NOT down the slope?

4. Originally Posted by anthmoo
Ohhh...hmm are you saying here that the accelleration is calculated in the x direction and NOT down the slope?
Careful!

Originally Posted by topsquark
When in doubt, try Newton's 2nd Law.
...
I am choosing a +x direction down the slope...
We can obviously find an acceleration component in the horizontal direction, but I can't think of any use for it.

-Dan

5. Originally Posted by topsquark
$w_x = w \cdot sin( \theta )$ <-- Make sure you understand why this is sine and not cosine!

-Dan
Wait...why is this sine and not cosine??

6. I'm going to try to do this in Paint. Many apologies, I need a new scanner!

-Dan

Edit: Well it's tiny, but you can see where the angle is. Remember we want the component of the weight DOWN the slope, which is the "opposite" side from the angle, so we use sine.

7. Originally Posted by topsquark
I'm going to try to do this in Paint. Many apologies, I need a new scanner!

-Dan

Edit: Well it's tiny, but you can see where the angle is. Remember we want the component of the weight DOWN the slope, which is the "opposite" side from the angle, so we use sine.

I'm sorry if I'm failing to understand too much but I always thought...

$sin(x)=\frac{opposite}{Hypotenuse}$

And down the slope appears to be the hypotenuse and the weight (acting directly downwards) seems to be the opposite, which gives:

$sin(\theta)=\frac{Weight}{Downslope}$

which gives

Am i missing something/doing something wrong??

EDIT: Or does the weight not extend to the base of the slope therefore we do not use that triangle?

8. Originally Posted by anthmoo
I'm sorry if I'm failing to understand too much but I always thought...

$sin(x)=\frac{opposite}{Hypotenuse}$

And down the slope appears to be the hypotenuse and the weight (acting directly downwards) seems to be the opposite, which gives:

$sin(\theta)=\frac{Weight}{Downslope}$

which gives

Am i missing something/doing something wrong??

EDIT: Or does the weight not extend to the base of the slope therefore we do not use that triangle?
Note where the right angle is in the component triangle for the weight. The hypotenuse is w and the two legs are wx and wy. wx is the leg "opposite" to the angle and wy is the leg "adjacent" to the angle. Try not to associate cosine with "x" and sine with "y" directions. It is often true that they are associated that way, but not all the time.

-Dan