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)
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)
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!