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# Where have I gone wrong?? [M1 Edexcel ]

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1. So I labelled the diagram as shown below:

α = 36.8698...

-30cos(36.8698...)-2(9.8)sin(36.8698...)=2a
-35.7600..=2a
a= -17.88 which is wrong??
Attached Images

So I labelled the diagram:

α = 36.8698...

-30cos(36.8698...)-2(9.8)sin(36.8698...)=2a
-35.7600..=2a
a= -17.88 which is wrong??
The 30N force is pointing to the right. So how can it have a component pointing down the slope?

A tip I find useful : Draw the 30N force as an arrow pointing away from P i.e. draw a horizontal arrow from P to the right and label it 30N.
3. (Original post by notnek)
The 30N force is pointing to the right. So how can it have a component pointing down the slope?

A tip I find useful : Draw the 30N force as an arrow pointing away from P i.e. draw a horizontal arrow from P to the right and label it 30N.
Am I not supposed to resolve it? As far as I know the forces should be parallel to the slope or perpendicular I guess.
4. You went wrong in life mah friend
Am I not supposed to resolve it? As far as I know the forces should be parallel to the slope or perpendicular I guess.
Your diagram would be correct if the 30N arrow was pointing to the left but it's not.

If you were to apply the two green forces to an object then it will result in a force to the left. Can you see that?

Try what I suggested: redraw the diagram with the 30N arrow pointing from P to the right. Then try and split the force into it's components.
6. Another thing, while it's not wrong to find a from tan a=3/4, they give you these numbers to make it easy to find sin a and cos a using a 3,4,5 triangle. sin a=3/5 cos a=4/5.
7. (Original post by notnek)
Your diagram would be correct if the 30N arrow was pointing to the left but it's not.

If you were to apply the two green forces to an object then it will result in a force to the left. Can you see that?

Try what I suggested: redraw the diagram with the 30N arrow pointing from P to the right. Then try and split the force into it's components.
So should it look like this?
Attached Images

So should it look like this?
I'm not certain but I believe 30sin a should be in the opposite direction
9. (Original post by miless090)
Another thing, while it's not wrong to find a from tan a=3/4, they give you these numbers to make it easy to find sin a and cos a using a 3,4,5 triangle. sin a=3/5 cos a=4/5.
Would you mean? Could you please elaborate?
10. (Original post by miless090)
I'm not certain but I believe 30sin a should be in the opposite direction
Really? Why's that?
So should it look like this?
No the two green forces would result in the blue force shown below:

If I draw the force like this then it may be easier to see what's going on:

The two components will be 1) up the slope and 2) downwards, in the same direction as the component of weight that is perpendicular to the slope.
Really? Why's that?
We were taught to create a triangle with the forces like this when resolving forces. If I'm honest I don't know why, but it's how we were taught and I've always got the right answer this way.
Attached Images

13. (Original post by miless090)
We were taught to create a triangle with the forces like this when resolving forces. If I'm honest I don't know why, but it's how we were taught and I've always got the right answer this way.
It's because if you have two forces as vectors F1 and F2 then the resultant force is equal to F1 + F2.

And if you think back to GCSE vector addition, you should be able to see why the triangle of forces works.
14. (Original post by miless090)
We were taught to create a triangle with the forces like this when resolving forces. If I'm honest I don't know why, but it's how we were taught and I've always got the right answer this way.
(Original post by notnek)
It's because if you have two forces as vectors F1 and F2 then the resultant force is equal to F1 + F2.

And if you think back to GCSE vector addition, you should be able to see why the triangle of forces works.
(Original post by notnek)
No the two green forces would result in the blue force shown below:

If I draw the force like this then it may be easier to see what's going on:

The two components will be 1) up the slope and 2) downwards, in the same direction as the component of weight that is perpendicular to the slope.
Oh I see now!! Thank you soo much!!
Would you mean? Could you please elaborate?
A well known right angled triangle is the 3 4 5 triangle. You're told tan a =3/4. At GCSE we were taught soh cah toa, tan = opposite/adjacent. So 3 is opposite the angle and 4 is adjacent to it. The hypotenuse is 5 so you can then find sin a (opposite/hypotenuse)=3/5 and cos a (adjacent/hypotenuse)=4/5
16. (Original post by miless090)
A well known right angled triangle is the 3 4 5 triangle. You're told tan a =3/4. At GCSE we were taught soh cah toa, tan = opposite/adjacent. So 3 is opposite the angle and 4 is adjacent to it. The hypotenuse is 5 so you can then find sin a (opposite/hypotenuse)=3/5 and cos a (adjacent/hypotenuse)=4/5
So how would you use this traingle in this situation, is it like the one you drew before, 30sina and 30cosa?
17. (Original post by notnek)
X
So where should I include the alpha angle for the orange arrows?

Attached Images

So how would you use this traingle in this situation, is it like the one you drew before, 30sina and 30cosa?
If sin a=3/5. 30sin a is just 30*3/5=18.
cos a=4/5. 30cos a is 30*4/5=24.
19. (Original post by miless090)
If sin a=3/5. 30sin a is just 30*3/5=18.
cos a=4/5. 30cos a is 30*4/5=24.
Oh,I see!! And does that work in all situations?
So where should I include the alpha angle for the orange arrows?

Your components should always be perpendicular. Your orange arrows are not but that may just be your drawing.

I recommend you draw your components like this:

Notice where I have put due to alternate/corresponding angles.

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