How tightly should a robot grip? The centre of mass matters too

The problem: the part rises, then tilts

A gripper picks up a metal bracket and lifts it; everything initially seems correct. Then the bracket tilts slightly and slips. Increasing the force might solve the problem, but could also deform a delicate surface. The useful question is therefore more precise than “how much does the part weigh?”: how much force does each contact need to support the part and prevent rotation?

This study builds an answer for a two-jaw gripper in a planar model. Following a synthetic one-kilogram bracket, we will see why a one-centimetre offset can require one-third more force, why friction is a limit rather than a force that is always present, and when squeezing harder stops being an admissible solution. This is a computed educational analysis, not a hardware test, a safety procedure or a detailed engineering design.

Before the equations: two different jobs

The jaws push horizontally against the sides of the part. This push is the normal force, because it is perpendicular to the contact surface. The part tends to fall: holding it requires vertical forces along the surfaces, called tangential forces. These are different directions. The normal force does not lift the part directly; through friction it enables a certain tangential force. Confusing them also leads to misreading a gripper’s force specification.

The second job is preventing rotation. A bag carried with two hands does not necessarily split its load equally: the location of its contents matters. This analogy helps visualise the problem but does not prove the result. For the bracket we will use two balances: forces determine whether it rises or falls, while moments determine whether it starts rotating. A moment measures the turning effect of a force through its lever arm, the perpendicular distance to its line of action.

A small model with visible assumptions

Imagine two point contacts at the same height, 60 millimetres apart: relative to the grip centre they are at x = −b and x = +b, with b = 0.03 metres. The centre of mass is horizontally offset by e = 0.01 metres to the right. The mass is m = 1 kilogram. The robot accelerates vertically upwards at a = 2 metres per second squared. We assume a rigid bracket, no rotation, no horizontal acceleration and no other contact.

Let N be the normal force of each jaw, equal on the two faces because of horizontal balance. It is not the sum of the two forces: the sum of their magnitudes is 2N. Let tL and tR denote the left and right tangential forces, positive upwards. A datasheet may use a different convention for gripping force: before comparing this calculation with a component, establish what that number measures. Every force result in newtons here refers to one jaw.

How much load must the contacts support?

At constant speed, balancing the weight mg would suffice. Upward acceleration also requires the force ma. We can combine them into an equivalent vertical load F. We have not increased the mass or introduced a mysterious extra force: we have rearranged Newton’s second law, tL + tR − mg = ma, moving the weight to the other side.

Here gravity g is 9.81 m/s². The result says that together the two contacts must exert 11.81 newtons upwards. It does not tell us how to divide them. Stopping here, we might choose tL = tR = 5.905 N and think the problem solved. That would be correct for a centred mass, but not for our bracket: the right contact must carry a larger share to prevent tipping.

The decisive step: balancing rotation too

Calculate moments about the centre of mass. The horizontal lever arms of the two contacts are −b−e and b−e. Because the contacts have the same height and the normal forces are equal and opposite, their moments cancel. With no angular acceleration, the following relation remains. Its second form uses the vertical balance already obtained: it immediately shows that a mass offset requires a difference between the tangential forces.

The ratio e/b is dimensionless: it compares the mass offset with half the opening. Here it is one-third. We obtain tR = 7.873333 N and tL = 3.936667 N. The right side carries twice the left; together they still give 11.81 N. The opposing moments are approximately 0.157467 N·m each: 7.873333 × 0.02 on the right and 3.936667 × 0.04 on the left. This check makes the uneven load distribution visible.

Available friction is not unlimited

To connect the vertical forces to the squeeze, we use a Coulomb static-friction model. At each contact the magnitude of the tangential force cannot exceed μN; μ is the dimensionless friction coefficient. Below this limit friction takes the value required by equilibrium: it is not always equal to μN. Equality marks the boundary of the sticking model, not a recommended operating margin. We assume the same μ on both sides.

Both contacts must satisfy the limit: the more heavily loaded one determines the requirement. The absolute value of e makes left and right symmetric; moving the mass to the other side changes the critical contact, not the minimum force. With μ = 0.4, divide 7.873333 N by 0.4 to obtain 19.683333 N per jaw. A centred mass would require 14.7625 N. The ratio is 4/3: the weight has not changed, but the rotational constraint requires one-third more squeeze.

This comparison exposes a common mistake: checking only 2μN ≥ F. With N = 15 N and μ = 0.4, total vertical capacity would be 12 N, apparently exceeding the required 11.81 N. Yet each side could supply at most 6 N, while the right needs 7.873333 N. Total capacity is not enough: it must be available at the right location. We cannot transfer spare capacity from the left to the right without changing the contact system.

Reading the plot: position and friction combine

The horizontal axis gives the mass offset in millimetres; the vertical axis gives minimum normal force per jaw. First notice the V shape: a centred position requires less force, and equal offsets to either side have the same cost. Then compare the curves: halving μ doubles the required force. The plot does not describe a probability of dropping the part; it describes the feasibility boundary of a model with fixed parameters. A curve below the dashed limit does not certify a real grasp.

Squeezing harder can become impossible

Now suppose an illustrative limit of 25 N per jaw. It might represent an actuator or part constraint, but we are not attributing it to a product. The condition becomes N_min ≤ 25 N. For our geometry this is equivalent to μ ≥ 0.314933. If actual friction were 0.2, the requirement would be 39.366667 N: asking the controller for a force beyond the limit does not solve the problem. The contact, grasp position, acceleration or mechanical support must change.

Another quantity that must not be confused with force is pressure. With a nominal area of 60 mm² per contact, 19.683333 N corresponds to a mean pressure of about 0.328056 MPa. The calculation is N divided by area, converting square millimetres to square metres first. This mean does not describe edge peaks, pad deformation or bracket strength. Doubling area halves mean pressure at fixed force; in the simple Coulomb model it does not automatically double friction capacity. These comparisons answer different questions.

Uncertainty: the best coefficient is not a design basis

So far μ, mass and position were known exactly. In a real process, contamination, material, wear, position and trajectory may change. To understand the effect without inventing a statistical distribution, consider hypothetical intervals: m ≤ 1.1 kg, a ≤ 3 m/s², |e| ≤ 15 mm and μ ≥ 0.25, keeping b = 30 mm. The formula increases with mass, upward acceleration and eccentricity, and decreases with μ. Its maximum over this rectangular set of intervals therefore occurs at the corresponding unfavourable extremes.

42.273 N greatly exceeds the illustrative 25 N limit. It is not an estimate of how many grasps will fail and not a universal safety factor: it is the deterministic result of less favourable assumptions. The intervals might be too wide, or omit still worse conditions. Their value is making assumptions open to examination and measurement: which surfaces really keep μ above 0.25? Does maximum acceleration include transients? Does the mass centre vary between batches? Without these answers, extra decimal places do not increase reliability.

Alternatives and boundaries of the result

Better centring reduces |e| without increasing compression. Lower acceleration reduces F but does not remove weight: even at a = 0, mg remains. A support beneath the part can transfer some load to another contact, while shaped jaws can change force geometry. These changes are not equivalent and cannot be assessed simply by inserting a more favourable μ into the same formula. Rebuild the balance with the new contacts and also examine access, deformation and release.

The model is planar and checks this specific load. Two frictional contacts in a plane do not demonstrate resistance to every three-dimensional disturbance: out-of-plane twisting requires a different analysis. We do not model pressure distributions, soft contacts, vibration, impacts, force-control errors or transitions between static and kinetic friction. Modern Robotics also distinguishes force closure, a geometric property, from actuator force limits. We are not proving a universal grasp: we are finding equilibrium for a stated case.

The formula also needs care outside the example: we assumed positive F. When |e| exceeds b, one computed tangential force becomes negative, directed downwards, forming a couple with the other. Absolute values formally preserve the limit in the ideal model, but contact feasibility, geometry and stability need re-examination. Extending a curve beyond the practical domain does not solve the problem. The published plots stay within |e| ≤ 20 mm, less than b.

What the calculations establish, and what remains to test

The attached program evaluates the balances and checks numerical consistency of force sums, moments and friction limits. It uses no random numbers, so no seed is needed; synthetic inputs are saved in the JSON. The short code below reproduces the main case, while the archive also contains alternatives and generation of the four localised plots. Its purpose is not to control a gripper, but to make every example number inspectable.

A real test would require measurement of actual force, contact characterisation under operating conditions and trajectory verification, with a protocol appropriate to the system. None of that was performed here. Industrial and collaborative robotics is a direction EL-AI intends to explore, as stated by the user; this study does not document a proprietary gripper, an installed cell or company experimental results.

Answer to the opening question

The gripper must squeeze enough for each contact to do its own job, not merely to support the total weight. Our ideal example requires at least 19.683333 N per jaw, compared with 14.7625 N for a centred mass: one centimetre of eccentricity costs one-third more force. If friction falls or constraints tighten, the solution can leave the admissible range. The design implication is concrete: before increasing the squeeze, understand where the load acts and which contact must carry it.

Sources and reproducibility

Primary teaching references: Kevin M. Lynch and Frank C. Park’s Modern Robotics materials, Chapter 12. The linked sections support the friction model and the distinction between force closure and actuator limits. The bracket geometry, parameters, derivation and calculations are this article’s educational construction; they do not reproduce an experiment by those authors and are not peer-reviewed original research.

Modern Robotics — 12.2.1 Friction.

Modern Robotics — 12.2.3 Force Closure.

m, g, a, b, e, mu = 1.0, 9.81, 2.0, 0.03, 0.01, 0.4
F = m * (g + a)
t_right = F * (1 + e / b) / 2
t_left = F - t_right
N = max(abs(t_left), abs(t_right)) / mu
print(f"{t_left=:.6f} N, {t_right=:.6f} N, {N=:.6f} N")

Code, data, and instructions · JSON. Educational calculations executed with Python 3.14.0; figures with Matplotlib 3.11.2. AI-assisted analysis, without claiming peer review or human review. Original illustrative ImageGen cover: it does not document EL-AI people, premises, or installations. Sources accessed 27 September 2026.