Friction on an Inclined Plane
Imagine a block placed on a ramp. If the ramp is flat, the block stays put. As you tilt the ramp, something changes. At a small tilt, the block still doesn't move — something is holding it in place. That something is static friction, acting up the slope, exactly balancing the component of gravity that tries to pull the block down.
Now tilt the ramp a little more. At some critical angle, the block just begins to slide. At that instant, static friction has reached its maximum possible value. Tilt it further, and the block accelerates downward — now kinetic friction acts up the slope, but it is weaker than the maximum static friction was.
That is the entire story in a nutshell. Let's make it precise.
The forces on an inclined plane
Take a block of mass m on a plane inclined at angle θ to the horizontal. The weight mg points straight down. Resolve it into two components:
- Perpendicular to the plane: mgcosθ — this pushes the block into the surface. The normal reaction N balances it: N=mgcosθ.
- Parallel to the plane (down the slope): mgsinθ — this is what tries to make the block slide.
Friction f acts parallel to the plane, opposing the relative motion (or the tendency to move). So if the block is at rest or sliding down, friction acts up the slope. If you push the block up the slope, friction acts down the slope.
Three cases
1. Block at rest (static friction)
The block does not move. So the net force along the slope is zero:
fs=mgsinθ
Here fs is static friction. It is not a fixed value — it adjusts itself to whatever is needed, up to a maximum. The maximum possible static friction is:
fs,max=μsN=μsmgcosθ
The block stays at rest as long as:
mgsinθ≤μsmgcosθ⇒tanθ≤μs
The angle θs=tan−1(μs) is called the angle of repose. At this angle, the block is just about to slide.
The angle of repose θs satisfies tanθs=μs. This is a direct link between friction and geometry — you can measure μs by finding the tilt at which sliding begins.
2. Block sliding down (kinetic friction)
Once θ>θs, the block accelerates down. Kinetic friction acts up the slope, with magnitude:
fk=μkN=μkmgcosθ
The net force down the slope is:
Fnet=mgsinθ−μkmgcosθ
So the acceleration down the plane is:
a=g(sinθ−μkcosθ)
| Condition | What happens |
|-----------|--------------|
| tanθ<μs | Block stays at rest |
| tanθ=μs | Block just about to slide |
| tanθ>μs | Block slides down with acceleration g(sinθ−μkcosθ) |
3. Block pushed up the slope …