A free-body diagram on a slope.

Step 1 of 5

Drawing the forces

A free-body diagram shows one body and every force on it, each force as an arrow leaving the body in its direction. This is how you build one:

1. Choose one body, here the block, and draw only that.
2. Add the weight: the Earth pulls the block straight down, without touching it.
3. Go over everything that touches the block. Here that is only the board, and it exerts two forces: a normal force pushing at right angles to the board, and friction along the board.
4. Do not add forces the block exerts on other bodies, nor “forces” that no body exerts.

Before you answer, try drawing the diagram on paper. Then choose On under Show the forces and compare.

Step 2 of 5

Splitting the weight

On a slope it helps to split every force into two components: along the board, and at right angles to it. Friction lies wholly along the board, and the normal force wholly at right angles to it. Only the weight, pointing straight down, is tilted relative to the board, so you split it, as you split vectors into components:

$W_{\parallel} = m g \sin\theta \qquad W_{\perp} = m g \cos\theta$$W_{\parallel} =$$m g \sin\theta \qquad W_{\perp} =$$m g \cos\theta$⁠

$\theta$⁠ is the slope angle. The component $W_{\parallel}$⁠ pulls the block down the board, and $W_{\perp}$⁠ presses it into the board. To remember which one has the sine, check a level board, $\theta = 0$⁠: there all the weight presses into the board ($\cos 0 = 1$⁠) and none of it pulls along ($\sin 0 = 0$⁠).

The block neither lifts off the board nor sinks into it, so at right angles to the board the net force is zero: $N = m g \cos\theta$⁠.

Now the board is tilted at $30^\circ$⁠, and the block is heavier: 2 kg. Work out the normal force, then check with Show the forces.

Step 3 of 5

The friction that holds

Along the board the net force is zero too, because the block stays put. So friction acts up the board, exactly as big as the component pulling down it. Work out how much friction acts on the block.

Step 4 of 5

A steeper slope

Show the forces is On, and the board is still at $30^\circ$⁠. Now the angle goes up to $38^\circ$⁠. The block will still hold, because the surface is rough. Predict what will happen to the normal force.

Try it: after you answer, change Slope angle between $0^\circ$⁠ and $38^\circ$⁠ and follow the arrows: the weight does not change, the normal force shrinks and the friction grows.

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Step 5 of 5

In summary

• A free-body diagram shows one body and every force on it: the weight, and each force from a body touching it. Leave out forces the body exerts.
• On a slope, split the weight into a component along the board, $m g \sin\theta$⁠, and one at right angles to it, $m g \cos\theta$⁠.
• The normal force balances the component at right angles: $N = m g \cos\theta$⁠. It equals the weight only on a level surface; on a slope it is smaller.
• A block resting on a slope: static friction equals the component pulling down the slope, $m g \sin\theta$⁠, and acts up the board.