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Mathematics 10 Online
OpenStudy (anonymous):

Physics;accounting for tension of a pulley on frictionless surface.

OpenStudy (anonymous):

OpenStudy (anonymous):

Tension of the pulley is calculated using........

OpenStudy (welshfella):

newtons laws of motion

OpenStudy (welshfella):

applying newtons law to the whole system force = mass * acceleration 6*g = 11 * a a = 6*9.8 /11 = 5.35 m s-2

OpenStudy (welshfella):

now i know how to find the tension in the string considering the horizontal string applying newtons second law;- Tension = 10 * 5.35 = 53. 5 N

OpenStudy (welshfella):

but you require tension in the pulley right?

OpenStudy (anonymous):

i think its mg= 6*9.8=tension on the string since the is no effect of 5 kg wt. as surface is frictionless

OpenStudy (welshfella):

sorry i dont know where i got 10 from!!!!

OpenStudy (welshfella):

if you consider the horizontal string its 5 * 5.35 not 10 * 5.35

OpenStudy (welshfella):

hhmm.. i'm not sure I would have thought you would have to take the 5 kg into account. Its been a long time since i did these...

OpenStudy (anonymous):

@welshfella how can you add 6+5= 11 to get 11*a this is not the case actually...

OpenStudy (welshfella):

but i'mm considering the whole system 5 kg and 6 kg are part of the whole system

OpenStudy (welshfella):

6 g is the downward force (dues to 6 kg mass) but 5kg is part of the mass

OpenStudy (welshfella):

the tension will be the same throughout the whole string so i guess thats also the force on the pulley

OpenStudy (anonymous):

5kg would have created friction on the surface due to wt. which would hav then created a resistance to allow the whole system to move down....this would hav decreased the tension on the string but here the surface is frictionless...

OpenStudy (welshfella):

hmm maybe but i dont think you can ignore the 5 kg...

OpenStudy (welshfella):

maybe i'm wrong

OpenStudy (welshfella):

I'd have to look it up

OpenStudy (anonymous):

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