Wednesday November 17th 2010

No class today!

Mason sent out some emails to people but i guess he wasn't feeling well enough to go through his contacts so here is the message:

Family is sick today. (My daughter brought home a fever and gave it to all of us.) Please chat with your project partners and lay out a plan for how you will complete the project in the remaining time. We will start harmonic motion on Friday. Please do the Pre-Lab on
"Measuring Spring Constants"

Please pass this on to other students.

Cheers,
mmason

So there you have it guys.

Remember, Due Friday:

Pre-Lab: Measuring Spring Constants

Lab Write Up: Problem # 3 Torques, Energy, and Friction (Monday's lab)

We should prepare for a "Fiesta" aka Quiz

We should be spending some time on our Projects and making sure we spread the work evenly over the next couple of weeks.

Friday November 12, 2010

Torque

Definition: Torque is the tendency of a force to cause or change rotational motion of a body. Torque is calculated by multiplying Force and distance, so the SI units of torque are newton-meters, or N*m

Torque is represented by the greek symbol τ (tau)

τ=r×Fsinθ

When the net torque =0, it means there is no acceleration or the object is moving with a constant speed.

When an object undergoes static equilibrium it must satisfy 6 conditions:
∑ F_ x=0
∑ F_ y=0
∑ F_ z=0
∑ τ_ x=0
∑ τ_ y=0
∑ τ_ z=0

Problem 1:
Find the ∑ τ about 20 cm of a meterstick whose mass is 160 g
τ=r×Fsinθ
τ=-(.3m)*(.16kg)*(9.8m/s^2)=-.48mN
0.3m=distance from the 20cm mark to the center of the meterstick (.50m-.2m=.3m) because it is an object with a uniform mass-the center of mass is at .5m

.16kg total mass of the ruler

9.8m/s^2=acceleration due to gravity

NEGATIVE SIGN=using the right hand rule, the thumb is pointing in or clockwise; therefore, it is negative.

How much mass do we have to add in order to get equilibrium?
0.2m×(0.160kg+x(extra mass))=0.8m×(0.16kg)
x(Extra mass)=.48kg

PROBLEM 2
A man weighs 50 N, is standing on a board with uniform mass that weighs 10N and has a lenght of 5m. The board is supported with two cords (one at each end of the board)
that make a 90degree angle with respect to the board.
The man is standing two meters away from one of the cords.

What is the tension of the two cords?
T1=tension of a cord
T2=tension of the other cord
WP= person's weight=50N
WB=board's weight=10N

∑ F_ x=0
∑ F_ y= T1+T2-WP-WB=0
T1=60-T2

∑ F_ z=0
∑ τ= T1*(5m)-WP*(3m)-WB*(2.5)=0
T1=35N

T2=T1-60
T2=35

These tensions were found by using one of the ends of the board as the pivot point. The location of the pivot point does not change the answer, it can only reduce or increment the number of unknowns and the number of possible equations.

HOMEWORK

Mastering Physics
Prelab(Forces, Torques, and Energy) page 96
Project #3 was assigned:Choose a problem from the physics book and make a video of it, step-by step procedure on how everything was done, 10 min Power Point Presentation. Groups of 2 people only, pick your partner.

Advanced Gravitation Notes

http://profmason.com/?p=1392