Where have I gone wrong?? [M1 Edexcel ]

Adorable98

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?? :frown:

(edited 7 years ago)

kl.png51.9KB

lab.png10.6KB

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Reply 1

Notnek

Original post by Adorable98

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?? :frown:

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.

Reply 2

Adorable98

Original post by notnek

Am I not supposed to resolve it? :s-smilie:

As far as I know the forces should be parallel to the slope or perpendicular I guess.

Reply 3

Son of God

You went wrong in life mah friend

Reply 4

Notnek

Original post by Adorable98

Am I not supposed to resolve it? :s-smilie:

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.

(edited 7 years ago)

Reply 5

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.

Reply 6

Adorable98

Original post by notnek

So should it look like this?

lab.png9.4KB

Reply 7

miless090

Original post by Adorable98

So should it look like this?

I'm not certain but I believe 30sin a should be in the opposite direction

Reply 8

Adorable98

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? :smile:

Reply 9

Adorable98

Original post by miless090

I'm not certain but I believe 30sin a should be in the opposite direction

Really? Why's that? :s-smilie:

Reply 10

Notnek

Original post by Adorable98

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.

(edited 7 years ago)

Reply 11

miless090

Original post by Adorable98

Really? Why's that? :s-smilie:

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.

kl.png53.0KB

Reply 12

Notnek

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.

(edited 7 years ago)

Reply 13

Adorable98

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

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!! :biggrin:

Reply 14

miless090

Original post by Adorable98

Would you mean? Could you please elaborate? :smile:

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

Reply 15

Adorable98

Original post by miless090

So how would you use this traingle in this situation, is it like the one you drew before, 30sina and 30cosa?

Reply 16

Adorable98

Original post by notnek

So where should I include the alpha angle for the orange arrows? :smile:

lab.png9.9KB

Reply 17

miless090

Original post by Adorable98

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.

Reply 18

Adorable98

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? :smile:

Reply 19

Notnek

Original post by Adorable98

So where should I include the alpha angle for the orange arrows? :smile:

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: