# Help With Understanding why weight in the x direction is MG*SIN*Theta

1. Oct 15, 2012

### Rabscuttle

This isn't a homework problem or anything, If I don't think about the formula that I use, and just plug in numbers I get the right answer. But I feel like a robot doing so, and it bothers me.

1. The problem statement, all variables and given/known data

On a typical problem where a wedge sits on an inclines plane, I am given the angle of the slope in degrees And the Mass of the object. I draw my right triangle and using the old convention of making the height of the right triangle equal to "O" (for opposite) and the base equal to "A" (for adjacent) and the hypotaneus. "H"

I set my Y axis along the normal. and my X perpendicular. I assume that now my hypotenous represents Weight in the X direction since my "O" is Weight.

The formula for the Weight in the X direction is MG*SIN(theta) as it says in the book.

Using methods I know I get MG/SIN(theta)

I know how to get MG*SIN(theta), by drawing my triangle elsewhere. I just have no idea why they are different.

I'm not sure if I am being unclear, I wish I could upload a diagram.

2. Oct 15, 2012

### howie8594

Where did you get MG/sinθ?

3. Oct 15, 2012

### Rabscuttle

#### Attached Files:

• ###### tringle.png
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4. Oct 16, 2012

### howie8594

Oh I see what you mean. Yeah I could see how anyone would be confused by that.

This is going to be hard to explain without a diagram, so try to follow along and let me know if you need clarification on anything.

You're just looking at the diagram wrong.

You have the diagram you drew right? It's much easier to understand if you draw a box of mass m on the slope.

Below that box, you can draw a line straight down representing mg. Now draw a line perpendicular to the slope below the box (perpendicular to the direction the box would slide). It's like drawing the line for the normal force but inside the triangle instead of outside it. The angle between the two lines you just drew will be the same as the angle of the slope with respect to the horizontal.

Now this might be a little confusing. Draw a line connecting those two other lines. That's the x-axis and it should be parallel to the slope. Yeah, the x-axis is not "A" in your diagram. It's actually the hypotenuse in your diagram. The y-axis by the way, is the normal force line perpendicular to the displacement of the box. There will be a 90° angle between the x and y axes. The hypotenuse should be straight down labeled mg.

Now you know mg, and you know the angle between mg and the y-axis (or normal force). From that, you can get the x component which again, should be parallel to the slope.

You should find that the x-component equals mgsinθ.

I know it's hard to understand this without a diagram so don't hesitate to ask stuff.

5. Oct 16, 2012

### lendav_rott

Try remembering it like this. The gravitational pull F = mg is made up of 2 components:
the slope directed component - mgsin θ
and the elastic force.
If you have a mass on a horizontal table, the mass pressures the table with its weight mg, but also the table is pressuring the block back by -mg - Newton's 3rd and 1st law essentially.

The thing is now, the elastic force between the block and the surface still exists, but the slope surface doesn't recognize it's being pressured by mg, a part of that mg is missing so to speak.

Imagine a right angled triangle - F = mg is the hypothenusis - the elastic force and the slope directed gravitational pull are the sides. Apply knowledge of Pythagoras himself and you will find what the components are.
To prove it you can apply the a² + b² = c², where the sides are mg sinθ and mg cosθ and the hypothenusis is mg.

Or by Newton's laws of motion - the 1st law states the a mass will stay still or will move with a constant speed without changing direction when the resultant force influencing the mass is 0. If the block is still on the slope at 1st and starts moving, then it has to gain the force somewhere otherwise you'd be contradicting Newton's laws.

6. Oct 16, 2012

### CWatters

If I've understood the question correctly this is the diagram needed...

#### Attached Files:

• ###### Forces on a box.jpg
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7. Oct 16, 2012

### howie8594

Yes, except the theta should be in between mg and mgcos(theta).

8. Oct 21, 2012

### Rabscuttle

After reading the replies I understand why I was wrong. Now it seems obvious to me :p. Thank you all for the diagram and explanation.