The Student Room Group

Maths A Level Friction Question

hi, i was kinda stuck on part b of this q. i got an answer of -0.815N for the tension, but would the magnitude of the force be twice as large as there is a force acting from the 0.8kg and 1.2kg object? or would it just be 0.815N? thanks.

i've also attached my working for the whole q.
(edited 7 months ago)
Reply 1
Original post by aditi_idk
hi, i was kinda stuck on part b of this q. i got an answer of -0.815N for the tension, but would the magnitude of the force be twice as large as there is a force acting from the 0.8kg and 1.2kg object? or would it just be 0.815N? thanks.

i've also attached my working for the whole q.

The tension/compression in the rod is just the magnitude of what youve calculated. As it acts in opposite directions on the two objects, if you summed them to get the resultant force, youd get zero. In fact, this is a good trick to calculate the acceleration of the overall system in part a) as a single equation which ignores the tension/compression as its an internal force (resultant force of the pair is zero) which is effectively the
2a = 2gsin(30) - ...
equation you get to near the end and you can write it down without explicitly modelling the individual systems with tension/compression.
(edited 7 months ago)
Reply 2
Original post by mqb2766
The tension/compression in the rod is just the magnitude of what youve calculated. As it acts in opposite directions on the two objects, if you summed them to get the resultant force, youd get zero. In fact, this is a good trick to calculate the acceleration of the overall system in part a) as a single equation which ignores the tension/compression as its an internal force (resultant force is zero) which is effectively the
2a = 2gsin(30) - ...
equation you get to near the end and you can write it down without explicitly modelling the individual systems with tension/compression.

thanks!
Reply 3
Original post by aditi_idk
thanks!


Without going into exactly what tension/compression is, you can think of it as simply a means of transmitting a force along the rod/string/wire/... and here the system could be modelled without it, but the two objects directly coupled (zero length rod). That way object A and object B exert equal and opposite forces on each other, a so called force pair. From a quick google
https://www.physicsclassroom.com/class/newtlaws/Lesson-4/Identifying-Action-and-Reaction-Force-Pairs
https://www.youtube.com/watch?v=Mk-PMZWcOcY&ab_channel=SiebertScience

As above, its probably slightly easier to answer part a) by considering the acceleration of the overall system A+B (so no need to model the internal tension). Then use one of the objects to calculate the tension/compression force acting on it to generate the calculated acceleration for b).

Quick Reply

Latest