- #1

Morsetlis

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I need... a lot of help. I spent an hour on this and a whole page of paper to no avail or certainty.

http://img88.imageshack.us/img88/8794/phuhelphd1.png

Suppose there is a diagonal downward force from top right to bottom left on an object with weight w on a surface with coefficient of friction u (static/kinetic friction aren't distinguished in this question.)

The diagonal downward force is a vector F_h with angle theta to the horizontal.

I have already figured out that, for at a certain angle theta, the force F required to push the object to overcome its frictional resistance, is

F_h = (uw) / (costheta - usintheta)

However, since every positive change in angle will reduce the horizontal component and increase the vertical component, this will increase the effect normal force on the object, which is N = u + F_hsintheta, thus also increasing the frictional force, F_f = uN.

At a certain angle, called the Critical Angle, F_hcostheta, which is the horizontal force required to move the object, will be equal to F_f, the frictional force opposing F_h. Increasing that angle will leave F_h < F_f and the object will not be able to move. After a certain interval of increasing degree, F_h will be greater than F_f once more, but the object will now move in the opposite direction.

Knowing that the Critical Angle forms a singularity at F_h = (uw) / (costheta - usintheta) so that F_h goes to infinity, I know I have to solve for (costheta - usintheta) = 0.

However, I also need to know the tangent of the Critical Angle, and this is not a happy answer, since my answer for the Critical Angle also included arcsin functions.

Here are my preliminary results:

Critical Angle = arcsin(- uw / (Fsqrt(u^2 + 1))) - arctan(-1/u)

tan(Critical Angle) = (Z + 1/u) / (1 - (-1/u)Z)

Z = -uw / (sqrt[ (F^2)(u^2 + 1) - (uw^2) ])

Z being a substitution since it would look horrible in the equation.

http://img88.imageshack.us/img88/8794/phuhelphd1.png

Suppose there is a diagonal downward force from top right to bottom left on an object with weight w on a surface with coefficient of friction u (static/kinetic friction aren't distinguished in this question.)

The diagonal downward force is a vector F_h with angle theta to the horizontal.

I have already figured out that, for at a certain angle theta, the force F required to push the object to overcome its frictional resistance, is

F_h = (uw) / (costheta - usintheta)

However, since every positive change in angle will reduce the horizontal component and increase the vertical component, this will increase the effect normal force on the object, which is N = u + F_hsintheta, thus also increasing the frictional force, F_f = uN.

At a certain angle, called the Critical Angle, F_hcostheta, which is the horizontal force required to move the object, will be equal to F_f, the frictional force opposing F_h. Increasing that angle will leave F_h < F_f and the object will not be able to move. After a certain interval of increasing degree, F_h will be greater than F_f once more, but the object will now move in the opposite direction.

Knowing that the Critical Angle forms a singularity at F_h = (uw) / (costheta - usintheta) so that F_h goes to infinity, I know I have to solve for (costheta - usintheta) = 0.

However, I also need to know the tangent of the Critical Angle, and this is not a happy answer, since my answer for the Critical Angle also included arcsin functions.

Here are my preliminary results:

Critical Angle = arcsin(- uw / (Fsqrt(u^2 + 1))) - arctan(-1/u)

tan(Critical Angle) = (Z + 1/u) / (1 - (-1/u)Z)

Z = -uw / (sqrt[ (F^2)(u^2 + 1) - (uw^2) ])

Z being a substitution since it would look horrible in the equation.

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