Static Equilibrium: Sliding and Tipping Bookcase

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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.
 
Attachments
  • Textbook figure 11.53: a rectangular bookcase on two pairs of legs, 2.00 m wide and 1.80 m tall, with a rope force F at angle theta applied to the upper corner; the 0.10 m leg offsets and the centre of gravity are marked.
    Physics Forums Pic.JPG
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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