Static Equilibrium: Sliding and Tipping Bookcase

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SUMMARY

The discussion focuses on calculating the force required to tip a bookcase, specifically addressing Part (c) of a homework problem. The user identifies the need to analyze forces and torques, utilizing equations for static equilibrium: sigma Fx=0, sigma Fy=0, and sigma tau=0. The user derives the tipping condition as F=w/(2sin(theta)), but recognizes the correct formula is F=w/((1/9)cos(theta) + 2sin(theta)). The challenge lies in incorporating the Fcos(theta) term correctly in the torque balance.

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  • Ability to analyze forces in a two-dimensional system
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The Head
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Homework Statement



See Picture: I am only stuck on Part (c) where it asks to calculate the force as the bookcase starts to tip.

Homework Equations



sigma Fx=0, sigma Fy=0, sigma tau=0

The Attempt at a Solution


I am setting up the problem by picking an axis that runs through the front two legs of the bookcase. The bookcase should tip if the horizontal component of the force is greater than the vertical component plus the weight, because then the vector sum of these forces points beyond the area of support.

The relevant torques that I see are:

Fsin(theta), Fcos(theta), weight (at point of cg), normal force of the back legs (but this is zero when the bookcase is about to tip).

Thus, for the point where tipping begins, I have:
(1.80 meters)*Fsin(theta)= 0.9m*(weight) + Fcos(theta)*0
I feel like I should use Fcos(theta) too, but runs right through my axis.

Thus I am left with: F=w/(2sin(theta))

The correct answer is F=w/((1/9)cos(theta) + 2sin(theta)), so I need to pick up a Fcos(theta)/9 somewhere but don't know how.
 

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The Head said:
Thus, for the point where tipping begins, I have:
(1.80 meters)*Fsin(theta)= 0.9m*(weight) + Fcos(theta)*0
I feel like I should use Fcos(theta) too, but runs right through my axis.
The axis is at the legs, and the vertical component of F runs 0.1 m away from it, along the side of the bookcase.

ehild
 

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