# Shouldn't constant velocity on an inclined plane be theoretically impossible given these assumptions?

For a scenario in which an object is sliding down an inclined plane, its motion opposed by a kinetic friction force, it seems to me that maintaining a constant velocity should be impossible given these assumptions:

1. The maximum coefficient of static friction is greater than the coefficient of kinetic friction.
2. When the object is at rest on the incline, the force of static friction is at its maximum value, and only a nudge is required to set it into motion. Thus, the angle of the incline in both cases (at rest and at constant velocity) should be (practically) the same.
3. There is no air resistance to be accounted for.

With this considered, wouldn't both scenarios yield the same value of tanθ (θ being the angle of the incline) for the coefficient of friction? Wouldn't this then break the assumption that max coefficient of static friction is greater than that of kinetic friction?

I've been puzzling over this because the whole "object sliding down an incline with constant speed" seems to be a common problem-solving scenario for introductory mechanics classes. However, with the assumptions I've listed above, I feel like this should be a theoretically impossible situation, unless, of course, one of the assumptions is wrong.

• I'm not sure I completely follow, but if a block is stationary on the ramp where the angle $\theta$ is such that static friction is maximal, then a small, brief nudge breaks static friction, and the block will start sliding. If the kinetic friction coefficient is smaller than the static one (which is typical), then the block will speed up down the ramp. The only way to get it to then move at constant speed would be to tilt down the ramp to a smaller $\theta$ (if that's possible in the setup). Can you clarify your understanding here? Commented Feb 8 at 16:51
• If the ramp is at an angle smaller than the critical angle to be at maximum static friction, then the block will need a larger nudge to get going. But, once moving, kinetic friction "takes over" (there is no more static friction because the block is sliding along the ramp), and if you happen to be at the critical angle for constant speed of sliding ($\mu_k = \tan\theta$), then the block will slide at constant speed. Do these comments answer your question? If so, I can write a brief answer. Otherwise, can you clarify where the issue might be> Commented Feb 8 at 16:53
• @march Actually, you're right. This would work at a smaller angle. I think the reason I wasn't considering that is because of the reliance on the formula saying that the coefficient of static friction equals friction force divided by normal force, when really that only applies when the frictional force is at a maximum. In the case of constant speed then, I guess 𝜇s would come out to some unspecified value greater than tanθ, while 𝜇k would still equal tanθ since it isn't bound to the same rule, which actually makes sense. Commented Feb 8 at 17:35
• When the object is at rest on the incline, the force of static friction is at its maximum value, - It does not have to be at its maximum value. Commented Feb 8 at 23:32

Static friction, which acts on a body at rest, applies just enough force to keep the body at rest, up to the maximum magnitude $$\mu_\mathrm{S}N$$ (frictional coefficient $$\mu_\mathrm{S}$$, normal force $$N$$). That is, the force could be anything from $$-\mu_\mathrm{S}N$$ to $$\mu_\mathrm{S}N$$, equal and opposite to any force that would otherwise dislodge the body.
Kinetic friction, which acts on a body in motion, always applies force $$\mu_\mathrm{K}N$$ (frictional coefficient $$\mu_\mathrm{K}$$) in the direction opposite the motion. If this force happens to be equal in magnitude to the gravitational acceleration $$F_\mathrm{G}$$ in the direction of motion, the body maintains constant speed.
In this way, one could have a body on a slope, held motionless by the action of $$\mu_\mathrm{S}(>\mu_\mathrm{K})$$ because the static friction exceeds $$F_\mathrm{G}$$ and thus supplies $$F_\mathrm{G}$$. Given a nudge, the body slides down the slope with constant speed (matching the initial nudge speed) because $$\mu_\mathrm{K}N=F_\mathrm{G}$$.