LocusBlog

Block sliding down a ramp

A block on a ramp is the first problem where you have to choose your own axes. Everything else in the question is easy. The whole difficulty is that gravity points straight down and the block does not move straight down.

The fix is to tilt the axes instead of the block. Put the $x$ axis along the slope, pointing downhill. Put the $y$ axis perpendicular to the slope. The block only moves along $x$, so the acceleration perpendicular to the slope is zero, and that fact is what gives you the normal force.

The free-body diagram, in words

Three forces act on a block sliding down an incline of angle $\theta$.

Weight $mg$ points straight down, toward the centre of the earth, not into the ramp. In tilted axes it splits into $mg\sin\theta$ down the slope and $mg\cos\theta$ into the slope. The angle between the weight vector and the perpendicular to the slope is the same $\theta$ as the ramp angle, which is the geometric fact the whole problem rests on.

The normal force $N$ points out of the surface, perpendicular to the slope. It is not equal to $mg$. It is whatever it needs to be to keep the block on the ramp, and since the perpendicular acceleration is zero, $N = mg\cos\theta$.

Kinetic friction points up the slope while the block slides down, because friction opposes relative sliding. Its magnitude is $f_k = \mu_k N$, which here is $\mu_k mg\cos\theta$.

Add the along-slope forces and divide by mass.

$$\begin{aligned} ma &= mg\sin\theta - \mu_k mg\cos\theta \\ a &= g(\sin\theta - \mu_k\cos\theta) \end{aligned}$$

The mass cancels. A one kilogram block and a hundred kilogram block on the same ramp with the same surfaces accelerate identically. Students distrust this result more than any other in the topic, and it comes straight from the fact that both the driving force and the friction force are proportional to $m$.

Frictionless, then rough

Set $\mu_k = 0$ and the formula collapses to $a = g\sin\theta$.

A 5 kg block on a frictionless ramp at $30^\circ$, with $g = 9.8$ m/s$^2$:

$$a = 9.8 \times \sin 30^\circ = 9.8 \times 0.5 = 4.9\ \text{m/s}^2$$

Half of $g$, because $\sin 30^\circ$ is exactly one half. The 5 kg never entered the arithmetic.

Now give the same ramp a kinetic friction coefficient of $\mu_k = 0.20$. $\sin 30^\circ = 0.5$ and $\cos 30^\circ = 0.866025$, so:

$$\begin{aligned} a &= 9.8\,(0.5 - 0.20 \times 0.866025) \\ &= 9.8\,(0.5 - 0.173205) \\ &= 9.8 \times 0.326795 = 3.20\ \text{m/s}^2 \end{aligned}$$

Friction removed about a third of the acceleration. Slide that block 4 m from rest and it arrives at $\sqrt{2ad} = \sqrt{2 \times 3.20 \times 4} = 5.06$ m/s instead of the 6.26 m/s the frictionless ramp would have given.

The bracket is also a brake. If $\mu_k\cos\theta$ exceeds $\sin\theta$ the bracket goes negative, and a negative answer here does not mean the block accelerates up the hill. It means a block that is already sliding down is slowing down. At $\theta = 20^\circ$ and $\mu_k = 0.35$ the bracket is $0.342 - 0.329 = 0.013$, barely positive, and a slightly larger coefficient flips the sign. The break-even is $\mu_k = \tan\theta$, which at $30^\circ$ is 0.577. OpenStax reaches the same relation by setting the acceleration to zero, which is the constant-velocity slide.

The sign mistake

Almost every wrong answer on this problem is one of two sign errors.

The first is swapping the sine and the cosine. The component along the slope is $mg\sin\theta$ and the component into the slope is $mg\cos\theta$. Check it against a limit. On a flat surface, $\theta = 0$, there should be no push along the ground and the full weight on the floor. Sine of zero is zero and cosine of zero is one, so $mg\sin\theta = 0$ and $N = mg$. Correct. Swap them and the flat floor launches the block sideways.

The second is adding friction instead of subtracting it. The formula is often written as $a = g(\sin\theta \pm \mu\cos\theta)$ in a formula sheet, and the sign is not decoration. Friction opposes the motion, so when the block slides down, friction acts up and the term subtracts. If you shove the block up the slope, it decelerates at $g(\sin\theta + \mu\cos\theta)$, because now gravity and friction both fight the motion and the two terms add. The same block on the same ramp has two different accelerations depending on which way it is moving.

A third mistake is not about signs. If the block is on the verge of moving rather than moving, $\mu_k$ is the wrong coefficient. Static friction is not a fixed value. It takes whatever magnitude the situation needs, up to a ceiling of $\mu_s N$, so $f_s = \mu_s N$ holds only at the instant of slipping. Below that the block sits still and the friction force is exactly $mg\sin\theta$, whatever $\mu_s$ happens to be.

Each simulation below integrates the motion from the forces rather than plugging into the formula, so you can vary the angle and the coefficient and watch the acceleration follow.

References