Three Forces in Equilibrium

An object is in equilibrium when the net force on it is zero. If two forces are applied and a third is required to balance them, the third force must be the exact negative of the sum of the first two — same magnitude, opposite direction.

The Condition

$\vec F_1 + \vec F_2 + \vec F_3 = \vec 0$

So:

$\vec F_3 = -(\vec F_1 + \vec F_2)$

This is just Newton's First Law expressed in vectors.

Step-by-Step Solution

Given:

• $\vec F_1 = 10\;\text{N}$ along the positive $x$-axis
• $\vec F_2 = 15\;\text{N}$ at $60°$ above the positive $x$-axis

Find: $\vec F_3$ for the system to be in equilibrium.

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Step 1 — Express each force in component form.

$\vec F_1 = (10, 0)\;\text{N}$

$\vec F_2 = (15\cos 60°,\; 15\sin 60°)$

$= (15 \times 0.5,\; 15 \times 0.8660)$

$= (7.500,\; 12.990)\;\text{N}$

Step 2 — Sum the two known forces.

$\vec F_1 + \vec F_2 = (10 + 7.500,\; 0 + 12.990)$

$= (17.500,\; 12.990)\;\text{N}$

Step 3 — The third force is the negative of this sum.

$\vec F_3 = -(17.500,\; 12.990) = (-17.500,\; -12.990)\;\text{N}$

Step 4 — Find the magnitude of $\vec F_3$.

$|\vec F_3| = \sqrt{(17.500)^{2} + (12.990)^{2}}$

$= \sqrt{306.25 + 168.74}$

$= \sqrt{474.99}$

$\approx 21.794\;\text{N}$

Step 5 — Find the direction of $\vec F_3$.

The vector $(-17.5, -12.99)$ is in the third quadrant. Computing $\arctan$:

$\theta = \arctan\!\left(\dfrac{-12.99}{-17.5}\right) + 180°$

$= \arctan(0.7423) + 180°$

$\approx 36.57° + 180°$

$\approx 216.57°$

(Equivalent to $-143.43°$ from the positive $x$-axis, or $36.57°$ below the negative $x$-axis.)

Step 6 — Verify the equilibrium.

$\sum F_x = 10 + 7.500 - 17.500 = 0 \;\;\checkmark$

$\sum F_y = 0 + 12.990 - 12.990 = 0 \;\;\checkmark$

The net force is zero in both components — equilibrium is confirmed.

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Answer: The third force needed for equilibrium is

$\boxed{\;\vec F_3 = (-17.500,\; -12.990)\;\text{N},\quad |\vec F_3| \approx 21.79\;\text{N},\quad \theta \approx 216.57°\;}$

This force points down-and-to-the-left, exactly opposite to the resultant of $\vec F_1$ and $\vec F_2$. When all three are drawn tip-to-tail, they form a closed triangle — that's the geometric signature of equilibrium.

Try It

• Adjust the magnitudes and angles of the two given forces.
• The third force (in green) automatically updates to maintain equilibrium.
• When all three vectors are drawn tip-to-tail (translate the third to the tip of the second), they form a closed triangle — that's the geometric way to see equilibrium.
• The HUD shows that $\sum F_x = \sum F_y = 0$ at all times.