Can this motor lift that arm? Torque answers the question.
Key idea
Torque is force times lever arm: $\tau = F \times d$ (newton-meters, N·m). To hold a mass $m$ straight out at distance $L$, a joint needs $\tau = m g L$, with $g \approx 9.8\ \text{m/s}^2$. Pick a motor with plenty more than that, often about twice, for speeding up and for friction.
Why it works
A servo's datasheet lists stall torque, the most it can give while not moving. Running near stall heats the motor and draws large currents. Designing for half of it keeps the robot cool and lets it move, not just hold.
Worked example: Hold a can
A 0.35 kg can held 0.20 m from the shoulder joint, arm weight ignored.
- $\tau = m g L = 0.35 \times 9.8 \times 0.20 \approx 0.69\ \text{N·m}$.
- Datasheets often use kg·cm: divide by 9.8 and multiply by 100: about 7 kg·cm.
Answer: About 0.69 N·m (7 kg·cm). Choose a servo rated well above that, around 14 kg·cm or more.
Worked example: Reach farther
The same can at 0.30 m.
- Torque grows in step with distance: $0.69 \times 1.5 \approx 1.03\ \text{N·m}$.
Answer: About 1.03 N·m.
Common mistake
Forgetting that holding a load farther out takes more torque, or using grams as if they were kilograms.