# Coefficient of friction between inclined plane and ground

Apply a vertical force to an inclined plane. What coefficient of friction is required between the bottom of the plane and ground to keep the system in equilibrium?

Referencing the image, the vertical force $$F$$ should subject the plane to some amount force in the horizontal direction. To maintain equilibrium, the coefficient of friction $$µ$$ must be great enough to satisfy:

$$0=F_x-µN=F_x-µF$$.

I am having trouble identifying $$F_x$$. If it were a sub-component force of $$F'_x$$ and $$F'_y$$ then the forces would cancel out. Ultimately, my evidence supporting the existence of $$F_x$$ is that an object sliding down the plane would move horizontally and should induce complimentary motion in the plane.

I am slightly ashamed to admit that I am not a student, but someone who shouldn't need to ask this question! I was hoping for an explicit response, as this is not a homework question. I have read the relevant force analysis of power screws, which yields the following equation when simplified by ignoring friction and using square thread. The question being answered in this analysis is, what torque is required to raise a load W using a power screw?

$$τ=R*F_{reaction}=R*W*\frac{\sin{a}}{\cos{a}}$$

Which should mean that $$F_X$$ in the inclined plane problem is: $$F_x=F*\frac{\sin{a}}{\cos{a}}$$. I can't figure out how to get to this answer and my reference does not show the steps.

• You seem to be asking two different problems here : a wedge with friction and a screw without friction. Please focus on one problem per question. Also, have you tried looking similar questions in this forum? eg Rotational Torque Versus Thrust on a Screw – sammy gerbil Jan 24 '20 at 1:04
• I can understand your point of view, but the question I am asking is "What is the required coefficient of friction between an inclined plane and ground to keep the system in equilibrium when a vertical downward force is applied to the plane?" The part I am having trouble with is identifying the magnitude of the horizontal component of this applied vertical force. To demonstrate that I have been examining this problem, I added the second example which requires a similar analysis of forces. In my own attempts, I end up with Fx = Fcos^2(a), whereas the likely answer is Fx = Fsin(a)/cos(a). – daDib Jan 28 '20 at 21:48
• Assuming there is no friction at the sloping face of the wedge, then the vertical component of the normal force $F_y'$ must balance the applied force $F$. So $F_y'\cos a=F$. The horizontal force $F_x$ is the horizontal component of $F_y'$. – sammy gerbil Jan 28 '20 at 21:59
• Your force diagram for the wedge is wrong. Understandably you assume that $F>F_y'$ but in this case it is the other way round. So you should be resolving $F_y'$ into vertical and horizontal components, not resolving $F$ parallel and perpendicular to the sloping face of the wedge. – sammy gerbil Jan 28 '20 at 22:02
• Could you go into some detail there, because that would surely be the reason I cannot get the correct answer. If the only applied force is $F$ in the vertical direction, then what would the value of $F'_y$ be? There is something fundamental that I have misunderstood. Thank you for your response. Edit: Is it impossible for a force to be applied directly to a slope as I have drawn? – daDib Jan 28 '20 at 22:11

Assuming that the sloping face of the plane is frictionless, any applied force $$P$$ must be normal to the surface, at an angle of $$a$$ with the vertical. (This is your force $$F_y'$$.)
You are told that the vertical component of $$P$$ has value $$F$$. The horizontal component of $$P$$ is the horizontal force $$F_x$$ which you seek. From the triangle of forces, $$\frac{F_x}{F}=\tan a$$
Where you went wrong is when you drew a triangle to resolve $$F$$ into components. Probably you assumed that the applied force could not have any horizontal component because you were not told about it. But there must be one to ensure that there is no applied force parallel to the surface.
• According to your answer, the assumption I made was that a force can be applied to a surface at an angle that is not perpendicular to that surface. I thought this kind of force application was shown in the typical 'block on an incline' problem where the vertical gravitational force is divided into components perpendicular to the incline and parallel to the incline. Are you saying that it is impossible to apply a vertical force to an inclined plane? That the full magnitude of $F$ must be directed perpendicular to the incline, in the direction of $F'_Y$? – daDib Jan 28 '20 at 23:40
• A vertical force on its own cannot be applied to the plane in this situation because the surface cannot provide an opposing force along the plane. Imagine you put a block with weight $F$ on this plane. It will slide down the plane unless you also apply a force with a component up the plane. – sammy gerbil Jan 28 '20 at 23:49