Minimum force to tip the bin over

Click For Summary
SUMMARY

The discussion centers on calculating the minimum force required to tip a bin over, specifically addressing the moment calculations involved. The weight (W) of the bin is established as 40g, and the participant seeks to determine the force (F) while considering the moments about the wheel. The correct approach involves accounting for both the horizontal and vertical components of force, as neglecting the vertical component leads to an incorrect conclusion. The discussion highlights a flaw in the problem statement from an A-Level Physics Examination, where assumptions about the handle's position relative to the wheel are not clearly stated.

PREREQUISITES
  • Understanding of moments and torque in physics
  • Knowledge of force components (horizontal and vertical)
  • Familiarity with A-Level Physics concepts
  • Ability to interpret diagrams and problem statements in physics
NEXT STEPS
  • Study the principles of moments in static equilibrium
  • Learn about force decomposition into horizontal and vertical components
  • Review A-Level Physics past papers for similar problems
  • Explore common pitfalls in physics problem-solving and diagram interpretation
USEFUL FOR

Students preparing for A-Level Physics examinations, educators teaching mechanics, and anyone interested in understanding the principles of static equilibrium and force analysis.

sdfsfasdfasf
Messages
75
Reaction score
12
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!
 
Physics news on Phys.org
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…
 

Similar threads

Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K