- #1
jonnyjonny
- 3
- 0
- Homework Statement
- The forces needed to consider for the tipping of a box
- Relevant Equations
- Moments
Forces
I would like to know the considerations needed to find the tipping of a box. It is given μk=0.2 and the box is 50kg
There are actually 2 ways of solving the problem:
1. To consider the moment about the centre of mass of the box, given that the normal force on the box will be redistributed.
So, considering moment about center of mass,
clockwise moment= Ff(0.5) + F(0.3)= 50(9.81)(μk)(0.5)+(600)(0.3)=229.05Nm
anticlockwise= (50)(9.81)(x) =490.5x Nm [Normal force x distance, x from the centre of mass]
For net moment about center of mass to = 0,
490.5x=229.05
∴x=0.467< 0.5 (Therefore, box won't tip because the normal force does not leave the box)
2. However, we can also consider the moment about the bottom right corner of the box, where it will potentially tip at.
In this case,
anticlockwise moment= W(0.5)= 245 Nm
clockwise moment= N(0.5-x) +600(0.8) = 496.18 Nm
∴ clockwise moment will result.
There seems to be a contradiction with the 2 methods as shown above. Could someone point out any errors or concepts that could have been missed?