Newton's Second Law: F = ma gave $F = ma$. When several forces act, which $F$? The answer is the net force, the sum of all the forces on your free-body diagram, with signs. When that sum is zero, there is no acceleration.
Key idea
$\Sigma F_x = ma_x$: choose a positive direction, add every force along that axis with its sign, then divide by the mass.
Why it works
Each force on an object contributes to its acceleration, and forces along a line simply add, with opposite directions subtracting. The object responds to the total, so only the net force sets the acceleration. A large push opposed by large friction can give a small acceleration.
- Draw the free-body diagram and choose a positive direction (usually the direction of the acceleration).
- Write $\Sigma F_x$: every force along the axis, with its sign.
- Set it equal to $ma_x$ and solve for the unknown.
- If forces act along $y$ too, do the same there, separately.
Worked examples
Worked example: Two opposite forces
A $5\ \text{kg}$ cart is pulled right with $30\ \text{N}$ and left with $12\ \text{N}$. Find its acceleration (right is $+$).
- $\Sigma F_x = 30 - 12 = 18\ \text{N}$.
- $a = \dfrac{18}{5} = 3.6\ \text{m/s}^2$ to the right.
Answer: $3.6\ \text{m/s}^2$
Worked example: Vertical forces
A $50\ \text{kg}$ model rocket has $800\ \text{N}$ of thrust. Find its acceleration.
- Up is $+$: $$\Sigma F = 800 - (50)(9.8) = 800 - 490 = 310\ \text{N}.$$
- $a = \dfrac{310}{50} = 6.2\ \text{m/s}^2$ upward.
Answer: $6.2\ \text{m/s}^2$ up
Worked example: Finding a force
What horizontal pull gives a $10\ \text{kg}$ box an acceleration of $1.5\ \text{m/s}^2$ against $20\ \text{N}$ of friction?
- $F - 20 = (10)(1.5) = 15$.
- $F = 35\ \text{N}$: $20\ \text{N}$ to cancel friction and $15\ \text{N}$ to accelerate.
Answer: $35\ \text{N}$
Common mistake
Using only the applied force in $F = ma$ and forgetting friction, weight or other forces. Check: list every arrow on the free-body diagram along the axis of motion, and make sure each one is in your sum with its sign.
Try it in the Science Lab: The Second Law: F = ma.