Resolving forces into components

What resolving means

• Resolving is the reverse of adding: taking one force and replacing it with two perpendicular forces that together have the same effect.
• The two replacements are called the components of the force.
• Resolving is useful because it lets you deal with each direction separately — the same trick used for projectiles.

$$F_x = F\cos\theta \qquad F_y = F\sin\theta$$

• $F$ — size of the original force (N)

• $\theta$ — angle between the force and the $x$-direction

• $F_x$ — component along the $x$-direction (adjacent to the angle → cosine)

• $F_y$ — component perpendicular to it (opposite the angle → sine)

• Draw the force as the hypotenuse of a right-angled triangle with $\theta$ at the tail. Then adjacent → cosine, opposite → sine. This works no matter how the angle is measured.

Pulling at an angle

• A rope pulled at an angle $\theta$ above the horizontal on a sledge has:
  • a horizontal component $F\cos\theta$ that does the useful pulling,
  • a vertical component $F\sin\theta$ that lifts, reducing the normal force.
• Because the normal force drops, the friction drops too — which is why pulling a heavy case at an angle can be easier than pushing it flat.
• Pushing down at an angle has the opposite effect: it increases the normal force and so increases friction.

Objects on a slope

For an object on a slope inclined at $\theta$ to the horizontal, the natural axes are along the slope and perpendicular to it. Only the weight needs resolving:

$$F_{\parallel} = mg\sin\theta \qquad F_{\perp} = mg\cos\theta$$

• $F_{\parallel}$ — component of weight down the slope, which is what makes the object slide

• $F_{\perp}$ — component into the slope, which the normal force balances, so $N = mg\cos\theta$

• $\theta$ — angle of the slope above the horizontal

• Sine goes with the slope angle here — a useful check is the extreme cases:

  • a flat surface ($\theta = 0°$): $mg\sin 0° = 0$, nothing pulls it along, and $N = mg$. Correct.
  • a vertical drop ($\theta = 90°$): $mg\sin 90° = mg$, it is in free fall, and $N = 0$. Correct.

• The steeper the slope, the larger the component down the slope and the smaller the normal force — which is why things slide more easily on steep slopes.

Deciding which components you need

• Choose axes that suit the problem: horizontal/vertical for most, along/perpendicular to the surface for slopes.
• Resolve every force that is not already along one of your axes.
• Add the components along each axis separately.
• Apply $F_{\text{net}} = ma$ to each axis on its own.