Thursday, November 19, 2015

CAPM Challenge

CAPM Challenge - Table Sliding

Find the acceleration of a ball rolling down a table using only a ball, chalk, and ruler.
Also, predict the velocity of the ball after 4 seconds.

To begin, Valentina and I set up our experiment. We raised a table up on wooden blocks to create a slant. Then, using a metronome, we marked the position of the ball at each half second as it rolled down the table.


We then measured the ball's travel from 0 seconds to 3.5 seconds. We took three trials of data and averaged all of the results. Them, we took these results and graphed the averages. 


Afterwards, we converted the centimeter measurements to meters so, when we test our data with the motion detector, it will be easier to compare. 



The equation for our line is X = 0.011t^2 + 0.0081
We know that acceleration is equal to the change in velocity/the change in time. Half the slope is equivalent to the acceleration. Since the slope of the line is equal to the velocity, we multiplied 0.011 by 2 and found the acceleration to be .022 m/s.

Then, to predict the rate of acceleration at 4 seconds, we used instantaneous velocity. 
We took our original equation X = 0.011t^2 + 0.0081 and plugged 5 in for t as well as 3.

0.011(5)^2 + 0.0081
= .2831

0.011(3)^2 + 0.0081
= .1071

Next, we subtracted the results at 5 from the results at 3. That number was then divided by two to find the instantaneous velocity of 4.

  (.2831 - .1071) / 2
= .088

We predict that the acceleration at 4 seconds will be .088 m/s

Thursday, November 12, 2015

Unit 2 Summary

Unit 2 Summary - Free Bodies and Trigonometry 

In this unit, our efforts were pushed forward by force! We learned about different kind of forces, how they react upon certain objects, and how to predict the force's measurement using trigonometry.

Force: An interaction between two objects
           A push or pull (Represented as Fpush/Fpull)
         
Forces speed up, slow down, stop, and change direction but only when forces are unbalanced.
If you are pushing or pulling on an object and it does not move, the push and friction will be equal.

Normal forces are the forces perpendicular to the surface of the object in analysis. Frictional forces are the forces parallel to the surface of the object in analysis. Gravitational forces pull the object in analysis down towards the Earth.

Some forces include:  Fg - Gravity: Pull between an object and the center of the Earth (weight)
                                    Fn - Normal: Crunching/Scrunching, a perpendicular force
                                    Ff - Friction: Parallel force, sharing force between two surfaces
                                    Ft - Tension: Force along a rope/chain/etc when taut/tight
                                    Fs - Spring: Returns to it's normal state when being compressed/stretched

To express these forces, we learned how to create a force of free body diagram. They display the forces acting upon an object at any given time.
To create a force diagram, you must:
1) Sketch the system and its surroundings
2) Enclose the system within a system boundary
3) Shrink the system to a point at the center of coordinate axes with one axis parallel to the direction of motion
4) Represent all relative forces (across the system boundary) with a labeled vertex
5) Indicate which forces (if any) are equal in magnitude to other forces

Using the example of a ball being propelled into the air out of a cart, we can observe several unusual rules of force.
The ball an cart move forward at a constant velocity. When the ball is thrown into the air, there is no other force acting upon it so the ball will continue moving forward at a constant velocity.
This is proved in Newton's Third Law: An object in motion stays in motion unless acted upon by an outside force.

The motion map of the ball will remain the same no matter how it is thrown because, once the ball has left the thrower's hand, gravity is the only force acting upon it.


We also learned how to calculate net force from both the horizontal and vertical directions. For the ball, the net forces will be set up in an equation. 
fVertical = fgravity > 0N
fHorizontal = 0N

Elaborating on this information, we learned how to create Free Body Diagrams at an angle. If an object is being pulled up by a force at an angle to it's surface, than the vectors representing the angled forces are also going to be angled on the diagram.

Since the tension force is angled upwards, component forces have to be added to balance the diagram. Tension gets divided into ftx and fty.

If the object is getting pulled with a force of 38N at an angle of 55o, we could find the force of friction using trigonometry.


The cos formula will give the length of ftx, which is equal to ff. 
Cos = adjacent/hypotenuse. We are trying to find the adjacent side, since ft is the hypotenuse. 

Cos 55 = adj/38N

38(cos 55) = .8408167379
This can be rounded to a friction force of .84N

All together, in unit 2, we learned about forces and how they affect various objects. We learned how to display actions of forces with Free Body Diagrams. We learned how to calculate the net force of an object and we learned how to find the amount of force projected on an object using trigonometry. 
                                      

Tuesday, November 10, 2015

Weight Challenge

The Weight Challenge

Determine the weight of the unknown object hanging from two objects at unequal angels to each other


My partner Valentina and I observed the setup and found that rope A was being pulled with a tension force of 0.9N at a 37-degree angle. Rope B was being pulled with a tension force of 2.25N at a 70-degree angle.

In order to find the weight of the box, we set up a free-body diagram. 

We found that the angle between ftBy and ftAx was 90o (Degrees). Then, we subtracted 37 from 90 and found the angle between ftBy and ftA equals 53.
With this technique, we also found the angle of ftBy/ftAy and ftB to be 20o.





With this new information, we decided to use trigonometry to calculate the weights of ftAy and ftBy.

Cos = adj/hyp

Cos 20 = adj/2.25n
2.25N(cos 20) 
= 2.11N

Cos 53 = adj/0.09N
0.09N(cos 53)
= .54N

Then, we just added together 2.11 and .54 to get a total of 2.65N. Since fg is equal to ftAy+ftBy, fg will also equal 2.65N. 

Therefore, the weight of the mystery box is 2.65N.