Static Equilibrium/Torque-Calculating the force required to tip a box

[taloz]

Static Equilibrium/Torque--Calculating the force required to tip a box

Homework Statement

A uniform crate with a mass of 13.2 kg rests on a floor with a coefficient of static friction equal to 0.536. The crate is a uniform cube with sides 1.21 m in length.
(a) What horizontal force applied to the top of the crate will initiate tipping?

(b) If the horizontal force is applied halfway to the top of the crate it will begin to slip before it tips. Explain.

Homework Equations

Fg= mg
Sum of torques=0
Sum of the forces=0

The Attempt at a Solution

If you apply a force to the top of the crate to initiate tipping the perpendicular distance between that force and the center of mass (my axis of rotation) should be 1.21/2 meters

The force of friction should also act 1.21/2 meters perpendicular of the center of mass (this is a cube after all).

Therefore, if the sum of the torques equals 0,

(1.21/2)*Force = (1.21/2)*Friction

(1.21/2)*Force = (1.21/2)*0.536*13.2*9.8

Force= .536*13.2*9.8


Webassign-- the thing i use to submit my homework online-- tells me that this is incorrect. Can anyone explain what I'm doing wrong to me. Thanks!

 

[taloz]

Nevermind, i see what the problem is. I'm forgetting to consider torque from the force normal. I'll have to change my pivot point to the area with the friction and Force normal and look at gravity and the force applied.