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.
That means the pendulum string goes up and to the right, at an 87 degree angle.
• 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.)
• What is the direction of the tension force exerted on the mass?
theta = tan^-1(2/.1) = 87deg.
• What therefore are the horizontal and vertical components of the tension?
horizontal component = 5N*cos(87) = .3N, vertical component = 5N*sin(87) = 5N
• What therefore is the weight of the pendulum, and what it its mass?
5N – 9.8m/s^2*mass = 0, therefore, mass = .5kg, 9.8m/s^2*.5kg = 4.9N = weight
Good, but:
The weight acts straight down, the y component of the tension acts straight up.
Assuming vertical equilibrium (i.e., no acceleration in the vertical direction, which since the pendulum is close to its equilibrium position is pretty much the case here), the weight is therefore equal and opposite to the y component of the tension.
So the weight is 5 N downward.
This agrees with your result, except for some roundoff discrepancy.
• What is its acceleration at this instant?
** **
20 mins
** **
&#Good responses. See my notes and let me know if you have questions. &#