http://www.xtremepapers.me/CIE/Internat ... 3_qp_4.pdf
Need help with Question 5 second part. Cant understand the direction of tension in the string.
Need help with Question 5 second part. Cant understand the direction of tension in the string.
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Maths (M1) may/june 2003
- Thread starter M@35TR0
- Start date
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- 40
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Hi There!!
Its pretty simple actually.
Basically according to the diagram tension in S1 would be due to both the weights of mass A and B
thus that is a total of 0.2+0.2=0.4g i.e 4N.
And as tension opposes weight that means that S1 has a tension of 4N upwards.
Similarly S2 has onl mass B acting on it thus the tention would be 2N
For the second part as the string S1 is cut tention of A is on string S2 so u need to apply Newtons law
T+ weight of A - air resistance = 0.2a
T+2-0.4=0.2a
For mass B tension is still upwards so:
Weight of B-Tension-Air resistance= 0.2a
2-T-0.2=0.2a
Solve simultaneously and u should end up with an answer!
Hope that helped!
Its pretty simple actually.
Basically according to the diagram tension in S1 would be due to both the weights of mass A and B
thus that is a total of 0.2+0.2=0.4g i.e 4N.
And as tension opposes weight that means that S1 has a tension of 4N upwards.
Similarly S2 has onl mass B acting on it thus the tention would be 2N
For the second part as the string S1 is cut tention of A is on string S2 so u need to apply Newtons law
T+ weight of A - air resistance = 0.2a
T+2-0.4=0.2a
For mass B tension is still upwards so:
Weight of B-Tension-Air resistance= 0.2a
2-T-0.2=0.2a
Solve simultaneously and u should end up with an answer!
Hope that helped!