a person is standing with there feet 0.1m apart and length from shoulders to feet(pivot point) is 1.5m what is the minimal coefficient of static friction required so that the frictional force is at least 1.5 x as large as the external force toppling the person?

according to my professor i have the following info
Fa= Wx 0.1m/1.5m
m=70kg
W=70x9.81m/s=687
687x(0.1m/1.5m)=48.5N
equation for static friction is
Force of Friction(Ffr)= mu x normal force (Fn)
but i simply do not know how approach the problem? please help

To find the minimum coefficient of static friction required, we need to set up equations for both the frictional force and the external force toppling the person.

Let's start with the external force toppling the person. The torque due to this force is given by the formula:

Torque = Force x Distance

Since the distance between the feet is 0.1m, we can rewrite the torque equation as:

Torque = Force x 0.1m

Since the person is in static equilibrium (not rotating), the sum of the torques around any pivot point must be zero. Taking the pivot point as the feet, we can write:

Torque due to external force - Torque due to friction = 0

Now, let's calculate the torque due to the external force. The force of the person's weight acting on each foot is given by:

Force of Weight = mass x acceleration due to gravity

Using the given mass of 70kg and acceleration due to gravity of 9.81m/s^2, we find:

Force of Weight = 70kg x 9.81m/s^2 = 687N

The torque due to the external force is then:

Torque due to external force = Force of Weight x Distance

Plugging in the values, we get:

Torque due to external force = 687N x 0.1m = 68.7Nm

Next, let's calculate the frictional force. The frictional force can be expressed as:

Frictional Force = Coefficient of Static Friction x Normal Force

In this case, the normal force is equal to the weight of the person. Therefore, the frictional force can be written as:

Frictional Force = Coefficient of Static Friction x Force of Weight

We want the frictional force to be at least 1.5 times larger than the external force toppling the person. Therefore, we can set up the following equation:

Frictional Force ≥ 1.5 x external force

Mu x Force of Weight ≥ 1.5 x Torque due to external force / Distance

Substituting the given values, we have:

Mu x 687N ≥ 1.5 x 68.7Nm / 0.1m

Simplifying the equation, we find:

Mu ≥ (1.5 x 68.7Nm / 0.1m) / 687N

Mu ≥ 15

Therefore, the minimum coefficient of static friction required is 15.