M1 1.1 - Modelling objects and forces

M1 1.1 - Modelling objects and forces

Choose useful mechanical models and explain what their assumptions allow you to ignore. A particle, a smooth surface and an inextensible string each simplify a different feature; using one does not silently give you the others.

Particles laminae and rigid bodies

Can the same suitcase be a particle while it slides across a floor, but need its size when it starts to topple? Yes: a model is chosen for the question, rather than attached permanently to the object.

To predict the suitcase's journey along a straight corridor, its dimensions may be unimportant. To predict toppling, the positions of the wheels and the edge matter. A model keeps the features needed for the prediction and neglects others.

Mechanical models

A particle has mass but negligible dimensions. Its rotation and the separate points at which forces act are ignored.

A lamina is a thin flat body whose thickness is negligible compared with its other dimensions. Its mass is spread over an area.

A rigid body does not deform: distances between its points remain fixed. A rigid body can still translate or rotate.

These descriptions answer different questions. A thin metal plate can be both a lamina and rigid. A particle is not massless, and a rigid body is not necessarily fixed in place.

Worked example 1: sliding luggage

Question. A suitcase slides along a straight corridor. Only the motion of its centre is required, and it does not topple. Explain why a particle model may be suitable.

Plan. Identify which physical features affect the required prediction.

The mass affects how the motion responds to forces. The suitcase's dimensions and rotation are not needed for this prediction.

Answer. Treat the suitcase as a particle: retain its mass and neglect its dimensions.

Check. This would cease to be suitable if the task required the suitcase's turning or toppling.

Change only the question asked about the same object. Size now becomes essential.

Worked example 2: toppling luggage

Question. The suitcase is tilted about one wheel. Explain why a particle model is unsuitable for deciding when it topples, and name a suitable model if its shape does not change.

Plan. Check whether separate positions on the object matter.

Toppling depends on where the weight acts relative to the wheel. A single point cannot represent those distances.

Answer. A rigid-body model is suitable; the suitcase's dimensions and possible rotation must be retained.

Check. “Rigid” is consistent with tilting: the body rotates without changing shape.

The useful simplification depends on the purpose.

Watch out

“Modelled as a particle” does not mean “no forces” or “no mass”. It removes dimensions, not the interactions causing motion.

Use the distinction in a new situation.

Light and uniform rods

A rod is long and thin, so its thickness can be ignored while its length is retained.

A metre rule and a rod with a heavy fitting at one end need different mass assumptions. Light means negligible mass in the model. Uniform means equal lengths have equal mass; for a straight uniform rod, its centre of mass is at its midpoint. A non-uniform rod has no such guaranteed midpoint.

Light and uniform

A light rod contributes negligible weight. A uniform rod has constant mass per unit length, so its weight can be represented by one force acting at its midpoint.

Uniform does not mean light. If a rod is non-uniform, use its stated centre of mass; do not assume it is at the midpoint.

For a straight rod of length LL, the midpoint is L/2L/2 from either end. The total weight still acts vertically downwards even if the rod is tilted. Calculating the turning effect of that force comes later.

Worked example 3: a uniform rod

Question. A uniform rod ABAB is 3.6 m3.6\,\mathrm m long and has weight 24 N24\,\mathrm N. State where its weight acts.

Plan. Use the midpoint guaranteed by uniform mass distribution.

AG=3.62=1.8 m.AG=\frac{3.6}{2}=1.8\,\mathrm m.

Answer. The 24 N24\,\mathrm N weight acts vertically downwards at GG, 1.8 m1.8\,\mathrm m from AA.

Check. GB=3.6−1.8=1.8 mGB=3.6-1.8=1.8\,\mathrm m, so the point is halfway along the rod.

Change only the mass distribution. The length alone no longer locates the weight.

Worked example 4: a non-uniform rod

Question. A non-uniform rod ABAB is 3.6 m3.6\,\mathrm m long and has weight 24 N24\,\mathrm N. Its centre of mass is 1.1 m1.1\,\mathrm m from AA. State where its weight acts.

Plan. Use the supplied centre of mass instead of the midpoint.

The weight acts at the stated point GG, so AG=1.1 mAG=1.1\,\mathrm m.

Answer. A vertically downward force of 24 N24\,\mathrm N acts 1.1 m1.1\,\mathrm m from AA.

Check. GB=3.6−1.1=2.5 mGB=3.6-1.1=2.5\,\mathrm m; unequal distances are allowed because the rod is non-uniform.

A small word changes the model materially.

Watch out

A light rod can transmit forces from attached objects. Neglect the rod's own weight, not every force acting on it. “Uniform” gives a midpoint centre of mass; it does not remove weight.

Check which information is actually supplied.

Strings pulleys beads wires and pegs

A string or wire constrains motion, and its stated properties tell you which forces and length changes can be ignored.

A string is flexible and pulls when taut; it cannot push. Its pulling force is tension, directed along the string away from the attached object. A light string has negligible mass. An inextensible string has fixed length: it does not stretch. This does not mean its tension is zero.

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