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.
** **
A simple pendulum has length 2 meters. It is pulled back 10 cm from its equilibrium position and released. The tension in the string is 5 Newtons.
Sketch the system with the pendulum mass at the origin and the x axis horizontal.
answer/question/discussion:
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 vector is 87.14 or 87 degrees counterclockwise from the positive x axis.
What is the direction of the tension force exerted on the mass?
answer/question/discussion:
??? The tension force is exerted by the pendulum string, so it is in the direction of 87 degrees also.
??? But we also have the force of gravity and whatever has managed to pull the pendulum back 10 cm from it’s equilibrium. That force could be horizontal, but I think it also could be at an angle, so I don’t know what direction it is.
Good, but the question asked only about the tension force, and you appear to have answered that correctly.
What therefore are the horizontal and vertical components of the tension?
answer/question/discussion:
X = T cos (87 degrees)
Y = T sin (87 degrees)
You were given that the tension is 5 Newtons. So you can easily find index and the y components by substituting this value into your correct equations.
What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion:
???
What is its acceleration at this instant?
answer/question/discussion:
???
** **
20 mins
** **
I believe that I am lost again. This is getting very frustrating.
Good description of the situation.
If you substitute the given tension you should have the information to answer the subsequent questions. Try doing so before you see my additional notes.
(The gravitational force is equal and opposite to the vertical component of the tension; there is nothing to counter the horizontal component of the tension, which will therefore be the net force. From the net force you can calculate the acceleration.)
Let me know if you have any questions about this; if so please go ahead and submit your revision. If you're sure of yourself, no need to submit the revision.)