- #1

anmsstu

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- Homework Statement
- The van is on an inclined plane (inclined by angle α) and has weight W. Wheels are distance 2d apart, COM is vertical distance h from inclined plane surface and parallel-to-plane distance d from both wheels and horizontal force acts through C. The van is stationary.

The first part of the question assumes both normal reaction forces on the wheels are equal and thus by taking the moment about COM the sum of frictional forces is zero.

The problem (second part of the question) asks to show for F such that van doesn't tip over :

W*tan(β-α) ≤ F ≤ W*tan(α+β) with tan(β)=d/h, W = mg

where normal reaction forces on wheels are not necessarily equal.

- Relevant Equations
- tan(β)=d/h

M = r×F

Okay so I'm not quite sure about the normal reaction forces. The condition is that they are both equal for the sum of frictional forces to be zero but then technically when substituting into the number equations in place of [ n(1) + n(2) ] either 2n(1) or 2n(2) makes me confused as then n(1) does not equal n(2) which would mean you cannot get rid of f(1) + f(2)? Unless f(1) + f(2) = 0 in all cases where van doesn't tip over. But then wouldn't that imply that if we have 2n(1) or 2n(2) then it will tip over? But this is the correct approach supposedly. Also I'm guessing 2n(`1) and 2n(2) are like the extremes of normal contact force at one wheel or the other which enable you to find the range for magnitude of F. I don't think I understand.