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Applying Newton’s Laws

•  CAPA due at 10pm tonight •  Will be a new CAPA

available – not due until Tues after the exam.

•  Clicker scores were updated and put on D2L.

•  Exam next Thursday at 7:30pm –  Calculators and 1 double sided

sheet of notes allowed –  Material up through today:

Chapters 1-5 Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

Announcements:

Clickers- scores thru 2/8 •  ClickersFeb Class Statistics •  Number of submitted scores:736 •  Average: 77.2 % •  Median: 87.2 % •  Standard Deviation: 23.3 % Look on D2L – if your clicker score is 0 or very

low, your clicker is not working or you are not attending. You also may not have registered your clicker or you are not properly registered in the class.

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9:00am Unregistered Clickers •  #0D40470A 47 #0E8D199A 108 •  #0F8C6AE9 93 #2651A9DE 79 •  #26657437 108 #26967ECE 106 •  #26A65BDB 87 #316EEBB4 68 •  #32AD801F 100 #333FB5B9 57 •  #3345CABC 101 #33506B08 71 •  #33536202 49 #33B10587 76 •  #33BC9817 79

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11:00am Unregistered Clickers •  #05C628EB 59 #25D76496 100 •  #261A132F 48 #266E420A 43 •  #267D90CB 93 #26C04FA9 57 •  #32829626 103 #330A0E37 84 •  #333FE9E5 103 #334081F2 40 •  #33B1EE6C 42

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Clicker question 1 Set frequency to BA

A.  T2 > T1 B.  T2 < T1 C.  T2 = T1

How do the magnitudes of the tensions in the rope compare?

M1 and M2 each must accelerate the SAME (they're tied together). The net force on M2 is M2*a. That's supplied by T2. So, T2 = M2*a. Now consider "M1+M2" as a system. That SYSTEM has mass M1+M2, and acceleration a. The only external force on it is T1. So T1 = (M1+M2)*a This is clearly bigger than M2*a. SO T1 > T2.

(T1 has to accelerate everything, T2 only has to accelerate M2)

Two blocks, with masses M2 > M1, are connected by ropes. You pull to the right on a second rope, with external force "T1". The blocks sit on a frictionless surface.

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Clicker question 2 Q. A mass m is pulled along a frictionless table by constant force Fext at some angle above the horizontal. Which statement below must be true?

Set frequency to BA

A.  N = mg B.  N < mg C.  N > mg D.  Impossible to tell

The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

Fext

Fext Free body diagram for the box

The vertical acceleration is zero since it is sliding along a table. Since Fextsinθ is positive,

A. Ffric > Fext , N > mg. B. Ffric < Fext , N < mg. C. Ffric > Fext , N < mg. D. Ffric < Fext , N > mg. E. None of these.

A mass m is pulled along a rough table at constant velocity with an external force Fext at some angle above the horizontal. Which statement below must be true?

Clicker question 3 Set frequency to BA

Fext Ffric

N

W=mg

m Fext

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Statics problems Statics problems are ones in which nothing is moving Thus, which we usually separate into components so and

Dubai Tower

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Statics problem

30 kg

A massless rope is looped over a pulley with one end connected to a 30 kg box and the other end held by a person. What force must the person exert to keep the box stationary?

Free body diagram for the box

Free body diagram for the hand .

In a massless rope the tension is the same everywhere!

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from the last problem we know

Statics problem

30 kg

What is the magnitude of the force exerted by the rod holding the pulley to the ceiling?

Free body diagram for the pulley

.

This is the force on the ceiling

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30 kg

The person holding the rope steps back a few feet. What is the magnitude of the force exerted by the rod on the ceiling now?

.

Clicker question 4 Set frequency to BA

A.  B.  C.  D.  E.  Impossible to tell

Pulley free body diagram

Frod < 2T and T=mg. Given the angles, we could calculate Frod.

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2. Apply and until you get the answer

Another statics problem

8 m 3 m

4 m

A person is suspending a 30 kg food bag 3 m in the air from two trees 8 m apart with 4 m high branches. How much force is needed to hold the (massless) rope in this position?

bag

W = mg

T T

1m 4m

This is more than twice what the bag weighs!

1. Draw free body diagrams for the relevant bodies. In this case, only the bag is relevant.

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Adding motion to the problem

25° m1

Two blocks connected as shown and no friction. If m1=10 kg and the blocks are moving at a constant velocity, what is m2?

Constant velocity means no acceleration so the net force (in any direction) is 0

Draw free body diagrams for both blocks

Block 1:

Block 2: Pick axes as and split the weight into components x

y

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An Atwood's machine is a pulley with two masses connected by a string as shown. The mass of object A, mA, is twice the mass of object B, mB. The tension T in the string is...

Clicker question 5 Set frequency to BA

A. B. C. Neither of these A B

A B

Since the weights are unequal, it should be clear that there will be some non-zero acceleration. Thus, since and are not zero, neither A nor B can be correct.

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A B

A B

We defined up as positive for both masses which means

& give

Full solution to Atwood’s machine

Solving for acceleration: so the tension is