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Moments question Mechanics June 2010

Hi all, was hoping someone could explain where I went wrong on this moments question on an OCR paper. The correct answer is 110N, I've attached both the question and my working

Thanks!

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Attachment not found
(edited 7 years ago)
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rsz_tsr_question_working.png343.2KB
Reply 1
Avatar for ak111
ak111
1
The weight is 45g.
You can work out the angle theta by using the 0.15m length they've given you at the bottom to form a right angled triangle with the 0.3m.
Then take moments about P.
Reply 2
Avatar for voltz
voltz
6
Original post by RBoss
Hi all, was hoping someone could explain where I went wrong on this moments question on an OCR paper. The correct answer is 110N, I've attached both the question and my working

Thanks!

Attachment not found
Attachment not found


So you've made a simple error which has complicated this.

First, looking at the definition of a moment:
"Force multiplied by the PERPENDICULAR distance from the pivot"

If you look at the diagram, the force F is perpendicular the the slab of length 0.600m, so we dont have to resolve any angle.

Simple, equal 0.600F to the clockwise moment and solve for F which will give you the answer of 110N.

Hope this helped:smile:
Reply 3
Avatar for RBoss
RBoss
OP
2
Original post by voltz
So you've made a simple error which has complicated this.

First, looking at the definition of a moment:
"Force multiplied by the PERPENDICULAR distance from the pivot"

If you look at the diagram, the force F is perpendicular the the slab of length 0.600m, so we dont have to resolve any angle.

Simple, equal 0.600F to the clockwise moment and solve for F which will give you the answer of 110N.

Hope this helped:smile:


I'm definitely overlooking something very obvious xD

I only formed the second equation cos theta = weight/F (using vertical resultant force = 0) in an attempt to eliminate the cos theta part from the moments equation, as the weight is not acting perpendicularly to the slab/slope. Surely only the normal component of the weight to the slope needs to be considered in the moments equation?
Original post by RBoss
I'm definitely overlooking something very obvious xD

I only formed the second equation cos theta = weight/F (using vertical resultant force = 0) in an attempt to eliminate the cos theta part from the moments equation, as the weight is not acting perpendicularly to the slab/slope. Surely only the normal component of the weight to the slope needs to be considered in the moments equation?


note: lengths on the exam paper diagram aren't to scale which could cause confusion.

You seem to have overlooked that the vertical component of F should be W/2... though you wrote it on line 1. Not sure if you've confused yourself by trying to make the vertical forces always equal W - but you don't want to do that when you're taking moments, There's a vertical component at the pivot but you can worry about that separately if you need to.

W=45*g = 441 N
cos ϴ = 1/2

fwiw I resolved F vertically and said

cos ϴ = F/Fvert
Fvert cos ϴ = F
W/2 cos ϴ = F

F=W/4
=110 N (3sf)

(there was no need to resolve horizontally for this question)

---
if would also work if you prefer to resolve parallel and perpendicular to the slab
Reply 5
Avatar for RBoss
RBoss
OP
2
Original post by Joinedup
note: lengths on the exam paper diagram aren't to scale which could cause confusion.

You seem to have overlooked that the vertical component of F should be W/2... though you wrote it on line 1. Not sure if you've confused yourself by trying to make the vertical forces always equal W - but you don't want to do that when you're taking moments, There's a vertical component at the pivot but you can worry about that separately if you need to.

W=45*g = 441 N
cos ϴ = 1/2

fwiw I resolved F vertically and said

cos ϴ = F/Fvert
Fvert cos ϴ = F
W/2 cos ϴ = F

F=W/4
=110 N (3sf)

(there was no need to resolve horizontally for this question)

---
if would also work if you prefer to resolve parallel and perpendicular to the slab


Aha, yes. I was looking too much at the parallel/perpendicular components, without considering Fv = W/2, as the gradient of the slope is the same, and F has twice the vertical distance of W... *facepalm*

That's a major bit of thinking space recovered though - thanks a lot for the clear explanation :biggrin:

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