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phy 231
Your 'cq_1_26.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_26.1_labelMessages.txt **
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Sketch a vector representing the direction of the pendulum string at this instant. As measured from a horizontal x axis, what is the direction of this vector? (Hint: The y component of this vector is practically the same as the length; you are given distance of the pullback in the x direction. So you know the x and y components of the vector.)
answer/question/discussion: ->->->->->->->->->->->-> :
The pendulum string direction is points below the positive x axis or down and to the right.
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What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
Tension force on the mass is up and to the left of the pendulum string, it has a positive y component, and a negative x component.
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What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
Theta from y axis at equilibrium = arc sin (0.10m / 4m) = 1.432 degrees
from the positive x axis the angle is 90 = 1.432 degrees = 91.4 degrees
horizontal component = Tension * sin (91.4 degrees) = 5N cos (91.4) = -.122161 N
vertical component = Tension * cos (91.4 degrees) = 5 N cos (91,4) = 4.998N
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What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
sum of forces in y = - m*g + 4.998N = 0
mass = 4.998N / 9.80 m/s^2 = 0.51 kg.
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What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
fx = mass * accel
accel = fx /mass = 0.122161 N / 0.51 kg = 0.24 m/s
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&#Very good responses. Let me know if you have questions. &#