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
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.
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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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.
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
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?
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
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