The Student Room Group

m2

Attachment not found
(edited 6 years ago)
Reply 1
Original post by Asad234
Attachment not found




need help pin all parts
Original post by Asad234
Attachment not found


"Attachment not found". What year and question number was this?
Reply 3
Original post by mupsman2312
"Attachment not found". What year and question number was this?



w.jpg
Original post by Asad234
w.jpg


Well, you'd have to use moments in two dimensions. It looks like a pretty tough question, though!
Reply 5
Original post by mupsman2312
Well, you'd have to use moments in two dimensions. It looks like a pretty tough question, though!


I know I have to use moments. How do I do it though?
Original post by Asad234
I know I have to use moments. How do I do it though?


Erm... :s-smilie:

I should know this - I did M2 last year! :colondollar:
Reply 7
Original post by mupsman2312
Erm... :s-smilie:

I should know this - I did M2 last year! :colondollar:


not helpful.
(edited 6 years ago)
Reply 8
Original post by Asad234
w.jpg

You've got two forces acting on the handbrake : F and T and any reaction forces from the pivot. Try taking moments about the pivot so you only need to consider F and T.

Post all your working if you get stuck.
Reply 9
Original post by Notnek
You've got two forces acting on the handbrake : F and T and any reaction forces from the pivot. Try taking moments about the pivot so you only need to consider F and T.

Post all your working if you get stuck.


I know you take moments from the pivot, but I don't know the angle for T

F x 350 x cos40= T x 60 x sin?

So just help me there.

If you know how to do it, please post the working instead, I'll be able to figure out my mistake myself. Thank you
Reply 10
Original post by Asad234
I know you take moments from the pivot, but I don't know the angle for T

F x 350 x cos40= T x 60 x sin?

So just help me there.

If you know how to do it, please post the working instead, I'll be able to figure out my mistake myself. Thank you



The perpendicular distance from the tension force to the pivot is the distance BP. You have angle CBP = 35 and BC = 60mm so you can find BP.

Then multiply the magnitude of T by BP to get the moment.

By the way you should really convert mm to m although it doesn't make a difference for this question.
(edited 6 years ago)
Reply 11
Original post by Notnek


The perpendicular distance from the tension force to the pivot is the distance BP. You have angle CBP = 35 and BC = 60mm so you can find BP.

Then multiply the magnitude of T by BP to get the moment.

By the way you should really convert mm to m although it doesn't make a difference for this question.


By your logic that would be wrong.

The answer to part two is 162N

I get this

F x cos 40 x 350= 1000 x 60 x cos 35

F = 183N

If you think I did it wrong, I am more than happy for you to correct me.
Reply 12
Original post by Asad234
By your logic that would be wrong.

The answer to part two is 162N

I get this

F x cos 40 x 350= 1000 x 60 x cos 35

F = 183N

If you think I did it wrong, I am more than happy for you to correct me.

cos 40 is incorrect. I'll give an explanation - give me a few mins.
Reply 13
Original post by Asad234
By your logic that would be wrong.

The answer to part two is 162N

I get this

F x cos 40 x 350= 1000 x 60 x cos 35

F = 183N

If you think I did it wrong, I am more than happy for you to correct me.

Notice that AB isn't horizontal so you can't just use cos(40).



Does this diagram help? If not, please explain how you calculate moments because I don't know which method you use.
Reply 14
r
Original post by Notnek
Notice that AB isn't horizontal so you can't just use cos(40).



Does this diagram help? If not, please explain how you calculate moments because I don't know which method you use.


How do you know that angle of F is 60?
Reply 15
Original post by Asad234
r

How do you know that angle of F is 60?


Wait, Now I know why.


Thanks, your method helped me.
I got 162
Reply 16
Original post by Asad234
r

How do you know that angle of F is 60?

The angle between BA and the hirizontal is 10 (alternate angles) so I marked 10 on the diagram. Then the angle between my red line and the vertical is a right-angle so you get 90 - 40 - 50 and I marked that on.
Original post by Asad234
Wait, Now I know why.

Thanks, your method helped me.
I got 162


Cool - that's what I got too. Sorry I couldn't help you out with this question - I think that I was just having a little bit of an off-day yesterday, and got scared by the look of the question... :redface:

Spoiler

(edited 6 years ago)

Quick Reply