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phy 201
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 **
Sketch the system with the pendulum mass at the origin and the x axis horizontal.
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Ok
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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.)
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I think it would be to the left
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What is the direction of the tension force exerted on the mass?
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upward against gravity
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What therefore are the horizontal and vertical components of the tension?
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2 cos(87) = .105
2 sin(87) = 1.997
I don't think this is right
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2 meters is the length. These would be the components of the length vector (or would had you included the units), provided the pendulum was pulled back 10 cm in the negative x direction.
The tension force on the pendulum mass makes the same angle as the string, since a string can only pull along its own line.
What therefore is the x component of the 5 Newton tension?
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What therefore is the weight of the pendulum, and what it its mass?
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Since I did not get the tension I dont think I can find it
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What is its acceleration at this instant?
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Accel would not be able to be gotten either without the others
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10 minutes
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You've got an angle, so you can find the components of that 5 Newton tension, and you should be able to work further into this problem.
Note that the 10 cm displacement is in the positive direction, so that the pendulum pulls up and to the left, not up and to the right. Your angle would therefore be 93 degrees, not 87 degrees.
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