Friction and critical angle question.

• Morsetlis
In summary: Z)Z = -uw / (sqrt[ (F^2)(u^2 + 1) - (uw^2) ])Z being a substitution since it would look horrible in the equation.In summary, the conversation discusses a problem involving a diagonal downward force on an object with weight w on a surface with coefficient of friction u. The solution involves finding the critical angle at which the horizontal force required to move the object is equal to the frictional force opposing it. This critical angle forms a singularity at F_h = (uw) / (costheta - usintheta) and can be solved for by setting (costheta - usintheta) equal to zero. However, finding the
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.

Last edited by a moderator:
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.

Morsetlis said:
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

Last edited:
Morsetlis said:
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?

Last edited by a moderator:
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...

1. What is friction?

Friction is a force that opposes the motion of objects when they come into contact with one another. It is caused by the roughness of surfaces and is dependent on the normal force and the coefficient of friction between the two surfaces.

2. How does friction affect the motion of objects?

Friction can cause objects to slow down or stop, depending on the direction of the applied force. It also causes objects to heat up due to the conversion of kinetic energy into thermal energy.

3. What is the critical angle of friction?

The critical angle of friction is the angle at which an object on an inclined surface will begin to slide. It is dependent on the coefficient of friction and the normal force acting on the object.

4. How is the critical angle of friction calculated?

The critical angle of friction can be calculated using the equation θ = tan-1(μ), where θ is the critical angle and μ is the coefficient of friction.

5. How does the type of surface affect the critical angle of friction?

The type of surface can affect the critical angle of friction by changing the coefficient of friction. Rougher surfaces will typically have a higher coefficient of friction, resulting in a lower critical angle, while smoother surfaces will have a lower coefficient of friction and a higher critical angle.

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