Static and kinetic friction.

Step 1 of 6

The limit of static friction

Static friction acts between surfaces that are not sliding over each other. It matches the force trying to move them, but only up to a limit that depends on the surfaces and the normal force:

$f_s \le \mu_s N$⁠

$\mu_s$⁠ is the coefficient of static friction, a number with no units that depends on the two surfaces: about $0.1$⁠ for ice, $0.4$⁠ for wood on wood and $0.8$⁠ for rubber. Surface grip in the Scene is $\mu_s$⁠.

On a slope the block holds as long as the component pulling down the board does not pass the limit: $m g \sin\theta \le \mu_s\, m g \cos\theta$$m g \sin\theta \le$$\mu_s\, m g \cos\theta$⁠. Divide both sides by $m g \cos\theta$⁠:

$\tan\theta \le \mu_s$$\tan\theta \le$$\mu_s$⁠

The block starts to slide at the angle whose tangent is $\mu_s$⁠. That is the red mark on the protractor.

The board is fixed at $30^\circ$⁠ and the surface is ice, so the block slides. Work out the smallest $\mu_s$⁠ that holds it in place, then raise Surface grip and check.

Step 2 of 6

A heavier block

The condition you found, $\tan\theta \le \mu_s$$\tan\theta \le$$\mu_s$⁠, has no mass in it at all: it cancelled. Now the board is at $21^\circ$⁠ and the surface is wood, $\mu_s = 0.4$⁠, and the 1 kg block holds. You will double the block’s mass. Predict what will happen.

Step 3 of 6

Kinetic friction

Once the block is sliding, the friction on it is kinetic friction. It no longer adjusts itself: it is constant, against the direction of sliding, and its size is

$f_k = \mu_k N$⁠

The coefficient of kinetic friction $\mu_k$⁠ is usually smaller than $\mu_s$⁠. That is why a heavy wardrobe is harder to start moving than to keep moving. In the Scene $\mu_k$⁠ is three quarters of $\mu_s$⁠: for wood, $\mu_k = 0.3$⁠.

Now the board is at $35^\circ$⁠ and the 1 kg block slides. Work out the friction force on it. Then choose On under Show the forces: the label of the arrow up the board now says kinetic friction.

Step 4 of 6

Down the slope

Along the board two forces act on the sliding block: the weight’s component down the slope, $m g \sin\theta$⁠, and kinetic friction up it, $\mu_k\, m g \cos\theta$⁠. The net force is the difference between them, and the second law gives the acceleration:

$a = \frac{m g \sin\theta - \mu_k\, m g \cos\theta}{m} = g\,(\sin\theta - \mu_k \cos\theta)$$a =$$\frac{m g \sin\theta - \mu_k\, m g \cos\theta}{m} =$$g\,(\sin\theta - \mu_k \cos\theta)$⁠

Here too the mass cancels. Work out the block’s acceleration down the board, at $35^\circ$⁠ and with $\mu_k = 0.3$⁠.

Step 5 of 6

Choosing an acceleration

Now the other way round: choose the acceleration and look for the slope. The board is at $20^\circ$⁠ and the block holds. Change Slope angle so that the block slides with an acceleration of about 2 m/s². The label of the green arrow shows the net force, and for a 1 kg block its number equals the acceleration.

Step 6 of 6

In summary

• Static friction matches the force trying to move a body, up to a limit: $f_s \le \mu_s N$⁠.
• On a slope a block starts to slide when $\tan\theta > \mu_s$⁠, whatever its mass.
• Kinetic friction acts against the sliding, and its size is constant: $f_k = \mu_k N$⁠, usually with $\mu_k < \mu_s$⁠.
• A block sliding down a slope accelerates at $a = g\,(\sin\theta - \mu_k \cos\theta)$$a =$$g\,(\sin\theta - \mu_k \cos\theta)$⁠, also independent of its mass.