## anonymous one year ago Physics;accounting for tension of a pulley on frictionless surface.

1. anonymous

2. anonymous

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

3. welshfella

newtons laws of motion

4. 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

5. 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

6. welshfella

but you require tension in the pulley right?

7. 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

8. welshfella

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

9. welshfella

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

10. 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...

11. anonymous

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

12. welshfella

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

13. welshfella

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

14. welshfella

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

15. 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...

16. welshfella

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

17. welshfella

maybe i'm wrong

18. welshfella

I'd have to look it up

19. anonymous

|dw:1439391345545:dw|

20. welshfella

21. anonymous

no effect of the 2 forces on each other due to surface is frictionless...

22. welshfella

if you check out the above link you'll see i'm right The 5kg has to be taken into account

23. welshfella

there is a tension between the 2 masses - the tension in the string

24. anonymous

may be you are right....i got confused...i hav to think again on it....

25. welshfella

its a tricky subject I wasn't sure but the link gives a good explanation.

26. anonymous

@welshfella yes your were right....i hav checked my book...my doubt is cleared....i am sorry that i wasted your time....but thanks....

27. welshfella

No you didn't waste my time - I had to make sure I was right also!!

28. welshfella

yw

29. anonymous

Hey guys thanks for all your insights

30. anonymous

This kind is one of the most counterintuitive

31. anonymous

Therefore so far to our best of knowledge tension 11kg*5.35m/^2

32. anonymous

I am crying right now

33. welshfella

yes - its s a bit hard to visualive - If the mass of the object on the table was say 15 kg there would still be motion because the table is frictionlass

34. anonymous

according to inertia that is.

35. anonymous

So it naturally falls upon us to calculate the acceleration and corresponding weight of the whole system

36. anonymous

Because tension is responsible for holding two masses which sum up to kg. Tension holds them from falling apart after all. Given that one side of the tension is pulling the system both masses are involved in the motion of the whole system.

37. welshfella

yes

38. anonymous

You established that "connection" with me here. Universal law of affection of physics applies to all of usXD

39. welshfella

we've found that the acceleration is 5.35 m s-2 but now we have to find the tension

40. anonymous

Isn't the tension 11kg*5.35m/s^2?

41. anonymous

most intuitively

42. anonymous

But tension on the pulley... I don't know

43. anonymous

that's the hardest part of this question is "tension on the pulley" instead of tension of the line.

44. welshfella

well the units of the tension will be N Newtons

45. welshfella

* will be in Newtons

46. anonymous

see what i hav checked in my book, (m1+m2)a = m2*g T= m2*g - m2*a

47. welshfella

yea that looks good

48. anonymous

T= m1*a

49. anonymous

Your equation assumes that overall force required to move the entire system is equivalent to the force exerted by a mass due to gravity.... Ok. This holds true on frictionless surface I suppose.

50. anonymous

*overall force required to accelerate

51. welshfella

so its 5 * 5.35 N?

52. anonymous

11*5.35m/s^2 assumes normal force is ignored?

53. welshfella

yes normal force acts perpendicular to motion so can be ignored

54. anonymous

No I think you should consider the "tension" which is induced by the 6kg mass as well. To assume there is only 5kg being responsible for the tension is absurd because it's not the only factor pulling the line.

55. anonymous

@welshfella got you .

56. anonymous

@Michele_Laino

57. anonymous

People say he's a physicist so

58. welshfella

can we apply newtons second law to the 5 kg mass in isolation? in which case we can say T = 5 * 5.35 = 26.75 N ?

59. welshfella

- why can't we do that?

60. welshfella

- that gives the tension in the horizontal string

61. anonymous

The prospect of assuming the tension to be "stretched" is somehow absurd?

62. anonymous

*5.35m/^2

63. anonymous

I think that reason why 6kg object is considered is because it's the source of force.

64. welshfella

I'll have to look at another video lol! I'm not saying you guys are wrong but i'm not convinced.

65. anonymous

@welshfella you are right T= 5 * 5.35 N

66. welshfella

I;ll give you a medal for saying that lol!

67. anonymous

So it's not the initial 11kg*5.35m/s^2

68. anonymous

but rather the second one.

69. welshfella

I'd like to giv medal to Robert also but it wont let me...

70. anonymous

that is final @welshfella is right...

71. anonymous

Thanks so much physics aficionados I think I understand it now.

72. anonymous

@Robert136 do you got it?

73. welshfella

yea the more i think about it.... the horizontal tension pulls the 5 kg mass so T must = mass * a = 5 * 5.35

74. anonymous

the more I think about it the better it settles in my brainXD

75. welshfella

but thats the tension in string what about the pulley? T acts directly on the pulley so that makes T on pulley = 26.75 N also?

76. anonymous

Right the tension is essentially moving the 5kg mass on frictionless surface so the tension must only account for the 5kg and greater acceleration implies proportionally valid force being applied. Ha! I think that's our ultimatum

77. anonymous

@welshfella your above statement is true...

78. welshfella

well guys we've learnt something with this session Good

79. anonymous

nice!

80. anonymous

The pulley is more like the center of this system at 90 degree angle. What if it was 180 degrere angle we all know to be the same?

81. anonymous

Since frictionless the direction must not matter ??

82. welshfella

right

83. anonymous

Mechanical advantage says it takes one half of force if pulley is used in two fold like manner. But I guess this doesn't concern this partiuclar problem...

84. welshfella

Its coming back to me now - after all these years - I wont say how many (lol!!)

85. anonymous

Physics is the protein bar for the intelligence

86. anonymous

Where intuition fails all the time. problem of this kind makes me reflect on my life from more logical poin of view.

87. anonymous

thanks all bye!

88. anonymous

Thanks everyone!

89. welshfella

the pulley must also be frictionless for the tension throughout the string to be uniform

90. welshfella

yw

91. anonymous

uniform acceleration that is

92. anonymous

Where veocity time graph would have a positive slope.

93. anonymous

Do you remember that stuff??

94. welshfella

some of it yes acceleration = rate of change of velocity = slope

95. welshfella

and distance traveled = area under the graph

96. welshfella

uniform velocity is indicated by a straight line with a positive slope

97. welshfella

* sorry that's uniform acceleration

98. anonymous

yeah whereas friction would prevent that

99. anonymous

One quick recap on this thread; when we did the calculation on acceleration of the whole system we used the sum of both masses.... is that accurate?

100. anonymous

As in 6.0kg*9.8m/s^2=11a as opposed to 5a just like what we did in the last counterpart.

101. anonymous

The object being acted upon is the 5.0kg mass accelerating... hmm my brain is such a pelletty equipment.

102. welshfella

yes we dealt with the whole system as we have no friction it was just like everything was horizontal with the force of gravity 6g acting on whole system and a weightless rigid body connecting the 2 masses

103. welshfella

|dw:1439398166592:dw|

104. welshfella

the double arrows indicate acceleration

105. welshfella

the force acting is 6g mass * acceleration due to gravity

106. anonymous

Ok. I get it. But how's that related to having only the 6kg object as being responsible for the entire system? I feel like 5.0kg is being "pulled" rather than moving "together" with the 6.0kg object. HmmmI guess I am really dumb

107. welshfella

the force has to accelerate the total mass which is 11 kg

108. welshfella

yes the 5 kg is being pulled by the tension in the string which is caused by the downward force of the 6 kg weight. I f there was no weight on the table the acceleration would be much greater . The 5 kg slows the whole system down. bit it is part of the system .

109. welshfella

If its weight was say 50 kg the system would still be in motion but acceleration would be a lot less.

110. welshfella

It is intuitive to a degree because the heavier the mass the acceleration would be a lot slower - because the force is constant at 6 g N but the mass is greater

111. welshfella

you can also find the tension in the string by considering the mass of 6 kg without referring to the 5 kg |dw:1439398934381:dw|

112. anonymous

Tension can be calculated with 6kg using the acceleratoin?

113. welshfella

the tension is working against the force 6g so the net force = 6g - T applying the second law 6*9.81 - T = 6 * 5.35 T = 26.76 N

114. welshfella

yes its valid to consider both masses separately

115. welshfella

in fact we could have found T and a by using 2 system of equations 6.g - T = 6 a - for the g kg mass T = 5a - for the 5 kg mass

116. welshfella

so we could have found T and a together

117. welshfella

adding we get 6g = 11a a = 6g/11 = 5.35 then T = 5 8 5.35 = 26.75

118. welshfella

* 5 * 5.35 = 26.75

119. welshfella

its all coming back now

120. anonymous

We all strive to find that momentXD Nostalgia old schoolXD

121. welshfella

yea lol!!

122. welshfella

Yea we had a very good math teacher. Our school was a very good one one - in Swansea UK. II guess its most famous pupil was Dylan Thomas the poet

123. anonymous

Awesome! I miss going to american schools:)

124. welshfella

Not that he was any good at Math. He was exceptional at English and cared nothing about anything else.

125. welshfella

yea At the time you didn't realize how good those days were.

126. welshfella

Have you ever heard recordings of Dylan Thomas reciting poetry? Some people don't like the way he reads them others find it awesome.

127. anonymous

Dylan Thomas don't gentle into that good night. Old age should rage at close of day. Rage rage rage into that good night.

128. welshfella

yea that's his most famous poem.

129. anonymous

It's so true though.

130. welshfella

Well id better feed the cat. She's in charge!! Yes. His father was the English teacher in the school and its to him the poem was written.

131. anonymous

Yeah every time I read his poem it makes me regret and realize so much I missed in life

132. welshfella

yep

133. anonymous

I guess I am supposed to be raging and raving but every day seems o be the reality of inevitable time passing.

134. anonymous

I wish I could have more emotions at work and assign a meaning to every day each day until I die. I guess travelling seems to be the best option for all of us.

135. welshfella

I think we waste a lot of time thinking about what we are going to do but never do it. Yea If you can afford it travel is great.

136. welshfella

nice talking to you must go

137. anonymous

Likewise! Thank so much for your help:) See you around.