Reason about forces and contact.

Touching is not the same as holding

Two jaws may reach an object without applying enough force to support it. The object’s mass sets its weight: mass in kilograms × 9.81 metres per second squared gives newtons. Friction at each contact must support part of that weight. An additional twisting moment can make the two required forces different.

The lab uses a planar object with contacts 0.1 metre to either side of its centre. Equal inward normal forces cancel horizontally. The vertical tangential forces must add to the weight, and their turning effects must cancel the external moment. These conditions describe required forces; they do not establish whether the contacts can supply them.

Position commands depend on compliance

A contact can compress. In the simple linear model, normal force equals stiffness × compression. With stiffness 2,000 N/m, compressing by 0.004 m produces 8 N. Millimetres must be divided by 1,000 before this multiplication.

A position command first closes the gap, then compresses the contact. The same displacement on a stiffer material produces greater force. Before contact there is no compression and no normal force; this contact cannot pull through empty space.

Force mode instead requests a normal force and calculates the displacement it would need. The lab assumes ideal static regulation, with a 20 N actuator limit and 12 mm travel per jaw. A soft contact or a large initial gap can use up travel before reaching the requested force. Requested and achieved quantities remain visible. This is a quasistatic comparison, not a simulation of feedback transients or impact.

Friction is a constraint, not a free resource

Static tangential force can take either sign, but its magnitude cannot exceed the friction coefficient times the inward normal force in this model. The planar friction cone shows those possible force pairs. Its width changes with friction; greater normal force increases the available tangential range.

If one required contact force lies outside its cone, the proposed static equilibrium is infeasible. At the boundary, the model has no friction margin. Passing with zero margin is not evidence that small disturbances or uncertain coefficients would be tolerated.

Check force and moment balance

Let W be weight, M the external counterclockwise moment and a = 0.1 m the half-distance between contacts. Upward contact forces L and R must satisfy L + R = W and aR − aL + M = 0. Therefore L = (W + M/a)/2 and R = (W − M/a)/2.

Moment is force times perpendicular distance and uses N m. One contact can require downward force when the moment is large enough. Both signs are allowed only within that contact’s friction capacity. The green wedge and required-force marks use the same inward-normal convention for each jaw.

Observe what an infeasible grasp does—and what is unresolved

For zero external moment, insufficient static friction gives a symmetric downward slip example. The short probe assumes a kinetic coefficient equal to 80% of the static coefficient, constant normal forces and continued contact. Subtract both upward kinetic-friction forces from weight and divide by mass to obtain downward acceleration. Starting from rest, displacement after time t is one-half acceleration × t squared.

The probe lasts 0.2 seconds. It omits jaw height, contact loss and deformation changes. With a nonzero external moment, the model does not invent a translational-only answer: rotation and changing contacts remain unresolved. A static failure must remain visible even when its subsequent trajectory is not simulated.

More squeeze has another cost

The fictional object has a declared 12 N per-jaw compression limit. A grip may satisfy friction requirements while exceeding that limit. The lab reports static feasibility and excessive compression separately before combining them for its narrow acceptance rule.

Holding this particular load is not a proof of force closure against arbitrary disturbances. It also does not establish dynamic stability, impact tolerance, material strength or hardware safety. Those questions need additional models and evidence.

Connect force to torque

A joint motor produces torque: a turning effort measured in N m. The Jacobian previously mapped joint speeds to endpoint velocity. Its transpose maps a requested endpoint force to the corresponding ideal joint torques. One way to understand this is conservation of instantaneous mechanical power: joint torque times joint speed must match endpoint force dotted with endpoint velocity in the ideal mechanism.

The lab supplies a fixed two-link example with explicit units. It excludes arm weight, inertia, motor gearing and torque limits. Mapping a force outside the contact cone into torques does not make that force feasible. Position control requests a configuration, force control requests an interaction force, and torque control requests joint effort; using one to achieve another needs a model and feedback.

Write a contact audit

Guided learners can compare a soft and stiff contact under the same closure, then repeat in force mode. Implementation learners should independently reconstruct the force/moment balance and check the Jacobian-transpose power identity. Decision learners should find a feasible non-damaging configuration and then test changes in friction, mass, gap and external moment.

Record which conditions failed and which dynamics were omitted. Temporary captures support this practice without awarding independently reviewed proficiency. Human review and learner pilots remain pending.