Why is tension not equal to weight in moments questions?

Even when it is said the system is in equilibrium and the object is stationary, in so many a level physics questions, the tension is not equal to weight. we know the system has to be in equilibrium becuase we are applying the Law of moments. Or, is the tension equal to the sum of the weight of different components in the system, that we are not given or able to calculate in the questions?

Reply 1

mqb2766

21

Original post by Nat4695

Even when it is said the system is in equilibrium and the object is stationary, in so many a level physics questions, the tension is not equal to weight. we know the system has to be in equilibrium becuase we are applying the Law of moments. Or, is the tension equal to the sum of the weight of different components in the system, that we are not given or able to calculate in the questions?

It depends on the question, but if the string was attached to the COM of the rod, then the tension would be equal to the weight. If you had a hinge at one end of the rod and a string at the other, then taking moments about the rod, the tension must be equal to half the weight to remain in equilibrium. If the string was close to the hinge, then the tension would be greater than the weight.

(edited 2 months ago)

Reply 2

Nat4695

OP

13

Do you mean the same as pivot when mentioning a hinge? Why is tension half the weight in the first scenario? Is it just to do with the moment created by the tension? as in the tension would be greater than the weight when it is closer to the hinge since perpendicular distance is smaller?

Reply 3

mqb2766

21

Original post by Nat4695

Do you mean the same as pivot when mentioning a hinge? Why is tension half the weight in the first scenario? Is it just to do with the moment created by the tension? as in the tension would be greater than the weight when it is closer to the hinge since perpendicular distance is smaller?

In a sense, a pivot or hinge is doing a similar thing and the tension not being equal to the weight (in general) is the mechanical advantage (usually) which is the Archimedes quote "give me a lever long enough and Ill move the world"
https://en.wikiquote.org/wiki/Archimedes

Tension is just a force so for the first scenario with a hinge at one end and a force at the other (tension) with the COM in the centre, then taking moments
2*tension = weight
due to the distances. If you put a long, light "rod" extension on the end so the COM remained where it was but the force was applied a long distance away, then the tension would be small (Archimedes). Obviously the hinge would have an upwards vertical component equal to the remainder of the weight (resolving vertically).

If the applied force was close to the hinge, then (taking moments) it must be >> weight so the hinge would have a downwards vertical component (resolving vertically).

If you think about the energy or work done (force*distance moved), then in the first case the force or tension moves a long distance (arc length) compared to the COM so the appllied force must be less than the weight for energy to be conserved. In the second scenario, the force must be > weight as it moves a smaller distance or arc length (smaller radius) compared to the COM. In both cases the COM and force/tension move through the same angular displacement, so the arc length (work done / conservation) is proportional to the radius or perpendicular distance (moments), so its effectively the same thing.