# Height of the force at which an object will tip

1. Nov 5, 2009

### vic226

1. The problem statement, all variables and given/known data
A bale is 0.25 m wide, 0.50 m high, and 0.80 m long, with mass 30.0 kg. The center of gravity of each bale is at its geometrical center. It is dragged along a horizontal surface with constant speed by a force $$\vec{F}$$. The coefficient of kinetic friction is 0.35.

A) Find the magnitude of the force $$\vec{F}$$.

B) Find the value of h at which the bale just begins to tip.

2. Relevant equations

$$\sum{\vec{F}} = 0$$
$$\sum{\vec{\tau}} = 0$$

3. The attempt at a solution

Part A is easy; for the bale to move at constant speed,

$$F = fr = \mu F_N = \mu mg = 102.9 N \approx 100 N$$

Part B is where I get stuck. I just randomly submitted the answer based on the picture's scale (I know it's not reliable but I had no choice, as it was a few minutes before due), and I got the answer h = 0.36 m. I know it has something to do with torque at equilibrium and the center of gravity of the system, but I fail to understand how it would work on such situation. Any pointers would be much appreciated.

2. Nov 6, 2009

### Andrew Mason

First of all the answer to part A is 103 N. to three significant figures.

The condition for falling is the torque about the centre of rotation - the leading bottom edge of the bale is forward. The centre of mass is behind the pivot point so there is a net backward torque. You have to find the point r such that

$$\vec F \times \vec r > m\vec g \times \vec r_{cm}$$

AM