Minimum force to tip the bin over

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Homework Help Overview

The discussion revolves around calculating the minimum force required to tip a bin over, focusing on the forces acting on the bin and the moments about the wheel. The problem involves understanding the roles of normal force, weight, and applied force in the context of rotational equilibrium.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the calculation of moments about the wheel, questioning the treatment of the vertical component of the applied force. There is discussion about the assumptions made in the problem regarding the positioning of the handle and its effect on the moments.

Discussion Status

Some participants have provided insights into the assumptions made in the problem, particularly regarding the vertical component of the force and its moment arm. There is an ongoing examination of the mark scheme and its implications, with differing opinions on its correctness and clarity.

Contextual Notes

Participants note that the problem is derived from an A-Level Physics Examination, raising concerns about the accuracy of the provided diagram and assumptions. There is a sense of frustration regarding the quality of the problem's presentation.

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Homework Statement
x
Relevant Equations
M = F x D
1713800647166.png

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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You are correct. Ignoring the vertical component is wrong based on the image and you would need the horizontal moment arm to account for it correctly.
 
1713803228888.png

This is the markscheme provided, does it provide any insight as to what the author has done?
 
Yes, the author is simply wrong.
 
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?
 
Most likely, but this is clearly not the case based on the picture and there is no mention of this assumption in the problem statement. That makes it wrong in my opinion. Particularly as an image is provided where this is not the case.
 
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:
 
Whoever constructed that clearly had a slip of mind and it should obviously not have passed quality control. Alternatively a case of the illustrator taking liberties to make the picture more realistic…
 

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