Minimum force to tip the bin over

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Homework Statement
x
Relevant Equations
M = F x D
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I am stuck on part b) i). I understand that there is no normal force from the ground as the bin is on the point of being lifted off the ground, that is all fine. That leaves R, F and W.

I know W is 40g, and I am required to calculate F, therefore it makes most sense to take moments about the wheel such that the force R is ignored. I want to find the force F such that the total moment is 0. I do not know the perpendicular distance from the line of action of F to the wheel, nor do I know the perpendicular distance from the line of action of the vertical component of F to the wheel. All I know is the perpendicular distance from the line of action of the horizontal component, which is clearly 1.3.

Are we meant to ignore the vertical component and just write 40g(0.3) = Fcos(20)(1.3)? This gives me the correct answer but I do not see why we should ignore the vertical component.

Thank you all so much!
 
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This is the markscheme provided, does it provide any insight as to what the author has done?
 
I suppose they have assumed that the handle is vertically above the wheel? Such that the line of action of the vertical component goes through the wheel itself and therefore doesn't provide any moment?
 
This is taken straight from an A-Level Physics Examination which is the highest level of examination in the UK you can take before university. :mad: