Tuesday, February 22, 2011
Movie Day!! :D
MythBusters video today about the myth that if one jumps right before your elevator plummets to its doom at the bottom of the shaft, they would live. Unfortunately it was busted and the dummy in the experiment gave a definitive splatter at the end of his ride. Using the equation V^2 = 2a * change in X (since the dummy's initial velocity is 0). The drop from the elevator is 92ft and gravity is an acceleration of 32ft/s^2, which would make the drop velocity 76.7ft/s. Unfortunately the average jump of a person is 8ft/s (using the same equation with a 1ft high jump), while this will slow your fall, it's only to a still bone crushing 68.7ft/s. The myth was busted.
Monday, February 21, 2011
Notes, February 17, 2011
Notes: February 17, 2011From the Force Diagrams and Net Force worksheet:
1. An elevator is moving up at a constant velocity of 2.5 m/s. The passenger has a mass of 85 kg.
a) Force diagram of the passenger:
b) Calculate the force the floor exerts on the passenger: (85 kg) x (10N / 1 kg) = 850 N
c) Calculate the force the floor exerts on the passenger:( 85 kg) x (10 N / 1 kg) = 850 N
5. A 70 kg skydiver jumps out of an airplane. Immediately after jumping, how large is the skydiver’s acceleration?
a) Acceleration due to earth’s gravity = 9.8 m/s ^2
b) Upon reaching a downwards velocity of 100 miles per hour, 300 N of drag resist the diver’s motion. Draw a force diagram and calculate how large is the skydiver’s acceleration.
Force unbalanced = (acceleration) x (mass)
Acceleration = (Force unbalanced) / (mass)
Acceleration = (700 N – 300 N) / 70 kg
Acceleration = 400 N / 70 kg
Acceleration = 5.71 m/s ^2
1. An elevator is moving up at a constant velocity of 2.5 m/s. The passenger has a mass of 85 kg.
a) Force diagram of the passenger:
b) Calculate the force the floor exerts on the passenger: (85 kg) x (10N / 1 kg) = 850 N
2. The elevator accelerates upwards at 2.0m/s^2
a) Force diagram of the passenger:
b) Equation for the vertical force on the passenger: Force unbalanced = (acceleration) x (mass)
c) Calculate the force the floor exerts on the passenger:Force unbalanced = (2.0 m/s ^2) x (85 kg)Force unbalanced = 170 NForce unbalanced + Normal force = vertical forceVertical force = 170 N + 850 NVertical force = 1020 N
a) Force diagram of the passenger:
b) Equation for the vertical force on the passenger: Force unbalanced = (acceleration) x (mass)
c) Calculate the force the floor exerts on the passenger:Force unbalanced = (2.0 m/s ^2) x (85 kg)Force unbalanced = 170 NForce unbalanced + Normal force = vertical forceVertical force = 170 N + 850 NVertical force = 1020 N
3. The elevator accelerates downward at 3.0m/s^2
a) Force diagram of the passenger:
b) Write the equation for the vertical forces on the passenger: Force unbalanced = (acceleration) x (mass)a) Force diagram of the passenger:
c) Calculate the force the floor exerts on the passenger:( 85 kg) x (10 N / 1 kg) = 850 N
4. While descending in the elevator, the cable breaks. How big is the force on the passenger by the floor?
a) At the moment that the cable breaks, the passenger and elevator accelerate downward at 9.8 m/s ^2.
b) Afterwards, the force of the floor on the passenger would be zero.
c) Relative to the elevator, the passenger would be weightless.
a) At the moment that the cable breaks, the passenger and elevator accelerate downward at 9.8 m/s ^2.
b) Afterwards, the force of the floor on the passenger would be zero.
c) Relative to the elevator, the passenger would be weightless.
5. A 70 kg skydiver jumps out of an airplane. Immediately after jumping, how large is the skydiver’s acceleration?
a) Acceleration due to earth’s gravity = 9.8 m/s ^2b) Upon reaching a downwards velocity of 100 miles per hour, 300 N of drag resist the diver’s motion. Draw a force diagram and calculate how large is the skydiver’s acceleration.
Force unbalanced = (acceleration) x (mass)
Acceleration = (Force unbalanced) / (mass)
Acceleration = (700 N – 300 N) / 70 kg
Acceleration = 400 N / 70 kg
Acceleration = 5.71 m/s ^2
6. A 900 kg car that exerts 5000 N of traction force on a level road while being opposed by 1000 N of friction and drag forces (combined)
a) Force diagram of car:
b) Write a net force equation for the car:
Force unbalanced = (acceleration) x (mass)
c) Calculate the acceleration of the car: Acceleration = (Force unbalanced) / (mass)
Acceleration = (5000 N – 1000 N) / (900 kg)
Acceleration = 4000 N / 900 kg
Acceleration = 4.44 m/s ^2
a) Force diagram of car:b) Write a net force equation for the car:
Force unbalanced = (acceleration) x (mass)
c) Calculate the acceleration of the car: Acceleration = (Force unbalanced) / (mass)
Acceleration = (5000 N – 1000 N) / (900 kg)
Acceleration = 4000 N / 900 kg
Acceleration = 4.44 m/s ^2
Friday Feburary 18
Today we discussed Atwood machines, or number 7 on the worksheet. For letter 'a', C has the greatest unbalanced force because more gravity is pulling down upon it. For letter 'b', we decided A has the least inertia because it has the least mass. Mr. Dickie decided that we do not know how to do letter 'c' because we did not do the lab that went with it.
Before we came to this conclusion, we talked about what a pulley does. It changes the direction of a force. We also came to the conclusion that an elevator, because of the counterweight in its cables, takes very little force to move.
Also, Mr. Dickie came up with a scenario where there were two blocks, where one was heavier than the other, and we were asked what the acceleration would be. We said it would be 3.3 m/s/s. After much discussion about how this is one system and using the acceleration equals unbalanced force/mass, or 10 N/3.0 kg, turns out we were right. We spent the entire class discussing this fact.
Before we came to this conclusion, we talked about what a pulley does. It changes the direction of a force. We also came to the conclusion that an elevator, because of the counterweight in its cables, takes very little force to move.
Also, Mr. Dickie came up with a scenario where there were two blocks, where one was heavier than the other, and we were asked what the acceleration would be. We said it would be 3.3 m/s/s. After much discussion about how this is one system and using the acceleration equals unbalanced force/mass, or 10 N/3.0 kg, turns out we were right. We spent the entire class discussing this fact.
Tuesday, February 15, 2011
Today we completed our elevator experiments by finding the acceleration of the elevator, how long the elevator accelerated for, and the distance that it traveled while accelerating. We found that the formula of _ a=F/m_ was not completely correct and changed it to
_ a=F(unbalanced)/m_. This means that if the upward force on you was 814 Newtons, and the downward force was 734 Newtons, your formula to find your acceleration would be
a=(814 N- 734 N)/ 73.4 kg
a=80 N/ 73.4 kg
a=1.09 (m/s)/s
We then discussed a few of the commonly missed questions from the test. We reviewed that, if there is no friction on the cart, than the forces are unbalanced, and therefore, if a constant force is applied, the cart constantly accelerates. Once the rope puling it breaks, there are only two forces acting on the cart which are balanced, and therefore the cart moves at a constant velocity.
Then we began talking about the force diagram for a balloon with a weight on the other end. We learned that the force pushing the balloon up is buoyancy from the air. There are two down arrows on the diagram, one for gravity and one from the tension from the string. When these forces are all balanced, then the balloon achieves equal buoyancy. If it is slightly pushed up or down, the forces are unbalanced, and it accelerates.
If the balloon, with the weight, is pushed toward the ground, the weight will touch the ground, and therefore be supported by it rather than the tension from the string. Because of this, the forces become, unbalanced, and there is less of a downward force, so the balloon accelerates upwards.
If the balloon is pushed laterally through the air, it eventually stops, because of the air resistance on the balloon.
_ a=F(unbalanced)/m_. This means that if the upward force on you was 814 Newtons, and the downward force was 734 Newtons, your formula to find your acceleration would be
a=(814 N- 734 N)/ 73.4 kg
a=80 N/ 73.4 kg
a=1.09 (m/s)/s
We then discussed a few of the commonly missed questions from the test. We reviewed that, if there is no friction on the cart, than the forces are unbalanced, and therefore, if a constant force is applied, the cart constantly accelerates. Once the rope puling it breaks, there are only two forces acting on the cart which are balanced, and therefore the cart moves at a constant velocity.
Then we began talking about the force diagram for a balloon with a weight on the other end. We learned that the force pushing the balloon up is buoyancy from the air. There are two down arrows on the diagram, one for gravity and one from the tension from the string. When these forces are all balanced, then the balloon achieves equal buoyancy. If it is slightly pushed up or down, the forces are unbalanced, and it accelerates.
If the balloon, with the weight, is pushed toward the ground, the weight will touch the ground, and therefore be supported by it rather than the tension from the string. Because of this, the forces become, unbalanced, and there is less of a downward force, so the balloon accelerates upwards.
If the balloon is pushed laterally through the air, it eventually stops, because of the air resistance on the balloon.
Thursday, February 3, 2011
Should Have Had a Snowday Day
On this wonderful, should-have-been-a-snowday, we white boarded the "Review Sheet" packet and managed to get through page 1 and 2. Seeing as this is a review packet, it is assumed that we already know this material thoroughly, so I will generally highlight the important points that will be important to remember for PART ONE of the test tomorrow.
-Unbalanced forces always lead to acceleration.
-Acceleration only occurs in the instant of the hit (see the bowling ball problem on page 2).
-Arrows can be added together ONLY if they have the same units. That is to say, do not EVER try to add together an arrow is in Newtons with an arrow in meters/second. Again, this point stems from the bowling ball problem on page 2.
-Newton's 3rd Law Force pairs must be the same type of force (like gravity) AND the agent-object is switched in the FBD's. An example of this would be:
-Unbalanced forces always lead to acceleration.
-Acceleration only occurs in the instant of the hit (see the bowling ball problem on page 2).
-Arrows can be added together ONLY if they have the same units. That is to say, do not EVER try to add together an arrow is in Newtons with an arrow in meters/second. Again, this point stems from the bowling ball problem on page 2.
-Newton's 3rd Law Force pairs must be the same type of force (like gravity) AND the agent-object is switched in the FBD's. An example of this would be:
Fg,earth-person
Fg,person-earth
-And yes, that too is another key point. You, as a person, are enacting a force of gravity on the earth just as the earth is enacting gravity on you, because gravity is a two-way street.
Also, because I promised it would appear in my Blogspot:
-BIRD IS THE WORD-
If you are having any trouble understanding concepts, I would find the Physics videos that Mr.Dickie has made; these can be found via Edline. They offer a straight-forward explanation of concepts, such as how to correctly use vector addition and component vectors.
Keep in mind that tomorrow, Friday, February 4, is the first part of a 2 part test; the second part is to be followed up on Monday.
Wednesday, February 2, 2011
Tsnownami Day
Well, apparently we did get hit with the Snopocalipse and have no class today. But we can still learn! Below are a couple videos I created to help explain forces and force diagrams. While watching them you should take notes and pause the videos to draw force diagrams and do calculations before I do them for you!
Tuesday, February 1, 2011
Today Mr. Dickie gave us a review sheet for us to go over for our test thursday.
Things on Worksheet:
Forces of two objects during a collision
Force diagrams
Newtons Third Law of Motion (An object at rest will stay at rest unless acted upon by an outside force) Be able to identify Newton's Third Law force pairs--opposite agent object. Ex. F a-x, F x-a (a and x don't represent anything in perticular)
Calculating force using Component addidion (making a triangle out of the free body diagram) and vector addition (lining the force vectors tip to tail in a triangle)
Things on Worksheet:
Forces of two objects during a collision
Force diagrams
Newtons Third Law of Motion (An object at rest will stay at rest unless acted upon by an outside force) Be able to identify Newton's Third Law force pairs--opposite agent object. Ex. F a-x, F x-a (a and x don't represent anything in perticular)
Calculating force using Component addidion (making a triangle out of the free body diagram) and vector addition (lining the force vectors tip to tail in a triangle)
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