Friction and critical angle question.

[Morsetlis]

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.

 

[OlderDan]

You did not state the whole problem. I think I managed to copy and paste your image and blow it up enough to read it. I will get back to you soon about your solution, but perhaps you could edit your post to make the problem clear to everyone.

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 [this cannot be right. μ is not a force. Maybe just a typographical error], 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. [Ah.. now I see this paragraph is part of the problem statement]

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.

I think you are making it much too hard. Set that denominator to zero as you did and solve the equality for the ratio sinθ/cosθ = tanθ. It is a very simple expression involving μ

 

[PhanthomJay]

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.

I think i have read the problem corectly, and it appears that you have correctly solved part a, and that your only question relates to part b. You again are correct that solving (costheta =usintheta) will give the critical angle. Try dividing both sides of the equation by costheta to see what you get. What is sintheta/costheta equal to in terms of the trig identities?

 

[Morsetlis]

I have already solved this problem, thank you for all your help.

I simply had to set costheta = usintheta and then found tantheta by taking tanarccotu = 1/u.

I was trying to do wave superimposition earlier...