Honors Physics 3rd hr. 1-31-2011

QUIZ FROM FRIDAY WAS PASSED BACK. TEST CORRECTIONS DUE WEDNESDAY.




Car Problem (from Friday)


Question: Find the mass of the car using both methods: vector addition and trig. ratios.

To the right is how you set up the force diagram to solve the problem with trig ratios.

First. we made a force diagram. The arrow pointing down is the force of gravity from the earth to the cart. The arrow pointing up to the left is the force of tension from the scale to the cart. And that is equal to 5 N. And the arrow pointing up and to the right is the normal force from the ramp to the cart.

The weight of the car will be equal to the arrow of gravity. Next, we drew the dotted lines to make triangles. And we were given the angle of the ramp was 30 degrees. And the tension arrow is equal to 5 N.

Now, we solved for ay.

sin 30 = ay/5N

sin 30 (5N) = ay

.5 (5N) = ay

ay = 2.5

Now that we found ay we can find a ax which is equal to bx.

cos 30 = ax /5N

cos 30 (5N) = ax

.866 (5N) = ax

ax = 4.33

bx = ax

bx = 4.33

Now that we figured out bx, we can find by.

tan 30 = 4.33 / by

tan 30 (by) = 4.33

by (.5774) = 4.33

by = 4.33 / .5774

by = 7.5

Now that we found ay and by we simply add them together to get the length of the gravity arrow which is also the weight of the cart.

ay = 2.5

by = 7.5

ay + by = Weight of the cart

2.5 + 7.5 = 10

Cart = 10 N

To find the weight of the cart simply convert Newtons to kg.

10 N = 1 kg.

Mass of the cart = 1 kg

Another way to figure out the problem is by using vector addition.

Rules of Vector Addition

• line up, tip - to - tail


• don't change the direction of the arrows


• 
  
  don't change the angle of the arrows


• don't change the magnatude of the arrows


• label all arrows


• remember you are adding arrows ( not numbers)

Since the forces are balanced, not necesarily equal, the arrows should start and end at the same spot. And you are given the same information. The angle of the ramp is 30 degrees and the length of the tension arrow is 5N. But, because you are using vector addition, it makes a 90 degree angle. And therefore the other angle is automatically 60 degrees because we know the angles of a triangle always add up to 180 degrees. If you do it that way, you can just use the properties of a 30, 60, 90 degree triangle and you will find out that the gravity arrow is 10N. And the mass of the cart is 1 kg. For vector addition, you can also use this method.

sin 30 = 5N / F g

Fg = 5 N / sin 30

F g = 5 N / .5

Fg = 10 N

Weight of cart = 10 N

Mass of cart = 1 kg

Those are two ways to figure out the car problem.

Today we had a quiz. After the quiz we did a problem similar to the car problem:

To solve this problem we draw a force diagram and solve using trigonometry.

Looking at the triangle, if we solve for gravity (G), we will be able to find the mass of the car.

sine62=5/G

G=5.66N

1/27/2011

Today in class, we opened by looking at a weight of 500g strung between two strings. Both strings were of different lengths, with the one on the left registering a force of 42 meters. Also, the alternate interior angle on the left hand side of the weight had a 64 degree angle. Thus, we had a right triangle on the left, with the longest side being 4 newtons, and the angle closest the weight, 64 degrees. Using sine, it could be determined that the missing side of that triangle opposite the 64 degree angle (Ay) had a force amount of 3.59 newtons. The we looked at the right triangle on the right hand side of the weight, opposite the one on the left.

Pause for a second, and draw a force diagram for the weight. There is the usual gravitational force, which in this case amounts to 5 newtons. Then draw the 2 lines to the right and left of the weight, these are Ax and Bx, and represent the bottom sides of the triangles. Using cosine, it can be determined that each side length (both Ax and Bx) have 1.75 newtons of force. Thus Ax equals Bx. Then, draw 2 arrows pointing upward, one representing side Ay and one representing By. Ay plus By = 5 newtons, the force of gravity. We know Ay, and we know the force of gravity, so by subtracting Ay from 5, we get 1.41 newtons. Then using the Pythagorean theorem with side Bx and By, it can easily be determined that the longer string, on the left has a force of 2.24 newtons.

Also, we learned that velocity is a vector.

Then we looked at a diagram of a boat traveling perpendicular to a river flow at 12 m/s, with the river flowing at 5 m/s. By forming a right triangle, with 5 and 12 as the legs, the speed of the boat as it travels at a sideways angle (along the hypotenuse) can be determined. In a similar way, when the boat is traveling in parallel to the current, with the river, it travels at 17 m/s and against, 7 m/s.

That was all we finished in class, before the bell rang as we were beginning to white board.

Post by Jordan Solano-Reed

1/26/11

Today in class we named the forces we already know: tension, gravity, the normal force, friction, and drag, which is the force from a fluid.

If a 5 kg string is hanging on a string connected to a ceiling, what is the tension in the string?
We drew a force diagram with the down arrow representing the Force of gravity from earth on the weight and the upward arrow of equal magnitude representing the Force of tension from the ceiling on the weight. The weight of the weight equals 50 N because the gravitational field strength equals 10 N/kg. Therefore, for every 1 kg of increase, there's an increase of 10 N. Since the earth is pulling down on the weight with a force of 50N, the tension in the string equals 50 N. Tension is the same everywhere in the string.

Strong Man problems:
If one horse weighs 800 N, the tension on the rope between the man and the horse equals 800N. The tension in the other rope the man is holding, whether another horse or a tree is attached to the other end, equals 800N. The tension of the rope between the man and two horses, weighing 800 N each, is 1600 N, and the tension of the rope between the man and the tree is 1600 N. If the 800 N horse is attached to the tree with a rope, the tension in the rope is still 800N.

Vectors in force diagrams:
You can make triangles out of a force diagram by using vector addition or vector components. Vector addition involves lining up the individual vectors tip to tail. This method is commutative, meaning that you can add the vectors in any order. To use vector components, draw dashed lines on the force diagram to make a right triangle using one vector and to represent the other vectors.
Just because a force diagram has three lines doesn't mean they can form a right triangle.

what we learned on 1/25/11

it takes more force to move two objects than 1 because you have the friction on the first object pushing it back which in turn puts more force on the second object because i has the the force of the first object and the force of friction the the second object

Class Notes for 1/24/11

Today in Honors Physics we began by white boarding problems #4 and #5 from the Free Particle Model Worksheet 2: Interactions. After discussing #4, we discussed a situation where Mr. Dickie is holding a bowling ball. Normally, we do not notice the effects of gravity while standing still. When on a swing, we feel the effects of gravity when we hit the ground. With just the bowling ball, earth is pulling on the bowling ball through gravity with 66 N. Mr. Dickie was holding the ball, thus exerting 66N of normal force on the ball in order to hold it up.

Overall (demonstrated by the force diagram on the right), the floor pushes up on both, so one arrow is used to represent it. The floor, through normal force, acts on Mr. Dickie with 792 N and 66 N on the bowling ball. Two arrows are used downward because in addition to the gravity of the earth on Mr. Dickie with 792 N, the normal force of the bowling ball acting on Mr. Dickie is 66N.

Class Notes for 1/21/11

We began the class by reviewing the forces we have studied in the past few weeks. Then, we labeled the various forces - gravity(G), air resistance(Air), friction(f), normal force (N), and tension (T). Next, we discussed how we can properly use these terms to label and identify forces in our force diagrams from now on. Finally, after looking at several examples we proved that friction forces are parallel to the surface an object is on, while normal forces are always perpendicular to the surface an object is on. At the end of class, we received a homework assignment regarding the forces involved in the interactions between two objects, and a review of the trigonometry involved in common physics problems.

Visualizing Forces with Springs

Today you'll be "seeing" the magnitude of forces acting on objects that are colliding. Before watching any of the videos you must make predictions as to which cart will experience the bigger force. The best way to express this is with free body diagrams.

For each situation you must draw a free body diagram for both carts during the collisions. So before you watch any videos, read the descriptions and make predictive FBDs.

Once you have your four sets of FBDs then download the videos and open them in Quick Time. Use the arrow keys to go through the videos frame by frame and observe the forces.

Find the videos at: http://fnoschese.posterous.com/