Play with itLoosen the grip until it slips
Worked solutionOne step at a time
Draw the forces on the box
Down: its weight, m·g. Sideways: each finger presses in with force N, and the two presses cancel. Up: friction on each of the two faces the fingers touch. Nothing else touches the box.
How much can friction carry?
Friction on one face can be anything up to μ·N, and no more. There are two faces.
Holding means friction can cover the weight
Solve for the squeeze
Answer: C, about 10 N
With μ = 0.5 each finger has to press as hard as the whole box weighs. Squeeze less and the box slides out, at an acceleration of g − 2μN/m.
What the simulation showedThe simulated box really does slide out
The reel's slip demo uses a heavier 1.5 kg block on purpose (so it does not give the quiz away), held by the Panda hand with friction set to exactly μ = 0.5:
- Theory says a 1.5 kg block (14.7 N) lets go when each finger squeezes less than 14.7 / (2 × 0.5) = 14.7 N.
- In the simulation, with the squeeze relaxing at 7.4 N per second, the block had crept 0.2 mm at 14.3 N and 1.5 mm at 13.9 N, then fell 13 cm in a third of a second. The small gap from 14.7 is just the block needing time to get moving.
- Squeezed at 40 N per finger, friction carried 14.7 N, not μN = 20 N. Friction only supplies what is needed, up to its limit. That is why the green arrows above stop growing once the box is held.
- The quiz box (1 kg) is held at 30 N per finger in the last scene, three times the minimum, and never moves while the arm sways.
Wrong answersWhy each one is tempting
| Pick | What went wrong |
|---|---|
| A · 2.5 | m·g·μ/2: multiplied by μ instead of dividing. |
| B · 5 | m·g/2: the friction each finger needs, not the squeeze that produces it. |
| D · 20 | m·g/μ: forgot there are two friction faces. |
Follow-ups an interviewer asks nextTry these before opening them
The robot now lifts the box upward at 5 m/s². Least squeeze?
Friction must supply m(g + a) instead of m·g: 1 × (9.81 + 5) / (2 × 0.5) = 14.8 N. Fast pick-and-place moves are where grasps actually fail, which is why controllers squeeze 2 to 3 times the static minimum.
The fingers get oily and μ drops to 0.2. Least squeeze?
9.81 / (2 × 0.2) = 24.5 N. Two and a half times as much for the same box. Gripper pads are rubber (μ near 1) for exactly this reason.
Why not always squeeze as hard as the gripper can?
Because the object has to survive it. An egg cracks at a few tens of newtons, a paper cup buckles, a ripe tomato bruises. The usable window is between m·g/(2μ) and the crush force, and a good gripper estimates μ or senses the first micro-slip to stay near the bottom of it.
The fingers hold the box off-centre, so it hangs to one side. What else must friction resist?
A torque. The weight now has a lever arm about the line between the two contacts, and the box will pivot between the fingers unless the contact patches can resist twisting. Point contacts cannot; wide, soft pads can. This is why pinch grasps go for the centre of mass.