1. Statics and Hanging objects
The sum of the forces is 0 (it is not accelerating)

2. Demo: Rope and bowling ball
1) Mass in center – pull on both ends to raise bowling ball –
Never get to perfectly horizontal rope!
Tension is the same for each side of the rope
2) Off set mass –
Which side has greater tension
Parallel with violin string on ring stands…
Pluck it –
The sum of the forces = 0 so we can set forces equivelant to each other and solve for unknown stuff!

3. The Hanging mass investigation – make a great diagram of your set-up diagram on board as well!
Set up a hanging mass like this: note force on each spring scale on both sides of the hanging mass. Note the hanging mass (mg). Note the different angles on both sides. + + - -

4. Your results use the diagram on the board during discussion
Are your results what you expected?
Was the force of tension the same on both sides when the angle was not the same on both sides?
What was the relationship between the force of the hanging mass (mg) and the force of tension on either side? (is Ft1 + Ft2 = Fmg?)
What are your calculated Ft1y, Ft2Y, Ft1x, Ft2x measures?
Does your calculated Ft1y + Ft2Y = Fmg? Should it?
Does your calculated Ft1x = Ft2x ? Should it?

5. A parallel situation to calculate and quantify the force of tension on both sides – Make a great diagram of this in your notebook! Diagram on board as well
Do not forget to put in your signs!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! +
Ft2
Ft1
ft1y
Ft2y - +
ft1x
ft2x
F mg -

6. The fun math part: the X component relationship
Sum Fx = 0 = Ft2x – Ft1x Therefore: Ft2x = Ft1x
Ft2 cos 27 = Ft1 cos 32
Ft2(.891) = Ft1 (.848) +
Ft2
Ft1
ft1y
Ft2y
32O
27O - +
ft1x
ft2x
F mg -

7. The fun math part – the Y component relationship
Sum Fy = 0 = Ft1y + Ft2y – Fmg so: Ft1y Ft2y = Fmg
Ft1 sin Ft2 sin = Fmg
Ft1 (.5299) + Ft2 (.454) = mg +
Ft2
Ft1
ft1y
Ft2y
32O
27o - +
ft1x
ft2x -
F mg

8. The fun math part – Put it together (substitution)
If you know Fmg (for this example we will use 10.N) you can solve for Ft1 and Ft2
Ft1(.848) = Ft2 (.891) (from x component relationship)
Rearrange: Ft1 = Ft2 (.891)/(.848) so: Ft1 = Ft2 ( )
and sub in to the Y equation +
Ft2
Ft1
ft1y
Ft2y
32O
27o - +
ft1x
ft2x -
F mg= 10.N

9. Like this - Ft2 = 9.89N Ft1 = Ft2 (1.05071) (from X relationship)
Ft1 (.5299) Ft2 (.454) = mg (from Y relationship)
Ft2 ( ) (.5299) Ft2 (.454) = 10. N
Ft2 (.55677) + Ft2 (.454) = 10N
Ft2 (1.011) = 10N
Ft2 = 9.89N

10. Finally solve for Ft1 Ft2(.891) = Ft1 (.848)
From the x component relationship we know that:
Ft2(.891) = Ft1 (.848)
Plug in your calculated Ft2 value and solve for Ft1
9.89 (.891) = Ft1 (.848) so Ft1 = 10.4 N
So we see that Ft1 > Ft2 and that is consistent with what we would expect to see considering the angle for 1 > the angle for 2 (like in the violin string demonstration)!!!!!!!!!!!!!!