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Phys_U3_Forces in 2-Dimensions

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Christian Wimber

on 6 January 2013

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Transcript of Phys_U3_Forces in 2-Dimensions

3. Students analyze objects affected by forces in two dimensions. 3.1 Students break a single force down into horizontal and vertical componants given its magnitude and direction.
3.2 Students find the net force acting upon objects by combining forces in the same dimension.
3.3 Students combine horizontal and vertical forces, providing both magnitude and direction.
3.4 Students, given a set of forces, can find the equilibrant force. Unit 3 Forces in 2-dimensions Targets Quiz 1 IQ's IQ 10.31 What was your hypothesis for the lab?
Did your data match your hypothesis?
If not, what is your new hypothesis?
In real life, if you pull on the strings as hard as possible, what angle will your angle be?
Does this match your new hypothesis? IQ 11.1 You are standing on an empty field. You take 15steps to the east and then 12 steps north.
Draw a picture of your path
How far are you from your original position?
If you were to walk straight back to your staring position, what angle do you make with the east/west axis? Notes Forces are Vectors Vectors can be added and subtracted, as shown below. A B C A+B=C A -B D A-B=D Example: Using what we have above, what is C+D? C D A A A single force can be represented by multiple vectors Force F F x F y x - used to represent horizontal forces
y - used to represent vertical forces Example: Split the force below into horizontal and vertical forces. Triginometry can be used to find the value of these new vectors. 15N 24° Soh Cah Toa see page 123 Examples: Find the horizontal and vertical componants for each of these forces. 68° 42N 24° 125N IQ 11.2 The sign to the left is hanging from a stretched cable. The tension on the cable is 535N. The cable is 42° above the horizontal. Draw a free body diagram for the sign.
Split each tension up into a horizontal and vertical force.
If the sign is (for the most part) at rest, what is Fnet on the sign? 3.1 Students break a single force down into horizontal and vertical componants given its magnitude and direction. 3.2 Students find the net force acting upon objects by combining forces in the same dimension. Just like other forces, we can split the net force up into horizontal and vertical. F F F F F F 1 1 1 2 2 2 x x y y Just as before, if Fnet =0, then acceleration = 0. Example:
For the sign, we know that...


because... What does Fg have to be? 48° F = 95N This lawn mower, when pushed as shown does not accelerate because of friction. Draw a free body diagram for the mower
What are a_x and a_y?
What are Fnet_x and Fnet_y?
Split the applied pusing force into its horizontal and vertical componants
What is the force of friction? 36.9° 53.1° 5.0N 5.0N 7.0N Is this object stationary?
Split all forces into horizontal and vertical
Find Fnet_x
Find Fnet_y 3.1 3.2 3.3 Students combine horizontal and vertical forces, providing both magnitude and direction. Once we have only Fnet_x and Fnet_y, we need to recombine them into an overall Fnet. Forces are vectors, so we need both magnitude and direction. Magnitude Use the pythagoreon theorum Example: Given the Fnet_x and Fnet_y below, calculate the magnitude of Fnet Direction Given Fnet_x and Fnet_y, we can use inverse tangent to find the angle of Fnet "opposite" "adjacent" Example: Calculate theta for Fnet as shown below: Example: For the net forces shown below, calculate both the magnitude and direction for Fnet (use the x-axis as a reference for the angle) 3.3 3.4 Students, given a set of forces, can find the equilibrant force. 640N 540N 36° 28° If mario wants to stop Bowser and Donkey Kong from moving the mushroom, what force will mario have to apply to the mushroom?
Give both magnitude and direction IQ 11.4 Please find Fnet_x and Fnet_y for the two forces below: 49N 38N 32° 28° Step 1: Split all forces in to Fx and Fy
Step 2: Find Fnet_x and Fnet_y
Step 3: Combine Fnet_x and Fnet_y
Step 4: Find the Equilibrant 540 x sin 28 = 253.5
640 x sin 36 = 376.2
-122.7

540 x cos 28 = 476.8
640 x cos 36 = 517.8
994.6

(-122.7^2 +994.6^2)^(1/2) = 1002 N
tan^-1 (-112.7/994.6) = -7.03°

1002 N, 7.03°, 1st quadrant IQ 11.8 In terms of F_1, F_2, theta_1 and theta_2...
find the horizontal and vertical components of each force
find the net horizontal and vertical forces
find the overall net force, including both force and direction Basically, I'm saying solve this problem with no numbers MQ 11.9 A 78kg skateboarder rides on a hill with an incline of 14° above horizontal. Find the portion of the skateboarder's weight that is perpendicular to the surface of the hill, and portion that is paralell. 14° 78kg Quiz 2 3.5 Students analyze forces on objects on inclined planes using rotated perspective.
3.6 Students analyze motion of objects on inclined planes. The equilibrant force is the force required to make acceleration equal to zero given any set of forces.
It is always the opposite of the net force. 3.4 3.5 Students analyze forces on objects on inclined planes using rotated perspective. We choose what we consider "horizontal" and "vertical" 78kg 14° Revisiting our skateboarder. Draw a free body diagram. Include friction, normal force, weight, and drag. This problem is solvable using what we already know but becomes significantly easier if we change perspective. Redifine:
horizontal is now paralell to the plane
vertical is now perpendicular to the plane 78kg 14° Example:
A 1.3kg mass sits at rest on a 25° inclined plane. Find the force of friction on the mass. 25° ES: 11.9 An 87kg man is climbing a Mountain with an incline of 52°. He slips and falls on ice, losing all frictional force. Calculate the tension on the rope. 52° THIS DOES NOT CHANGE THE DIRECTION OF ANY FORCE! IQ 11.10 A 3.50kg mass rests on a frictionless plane with a 10.5° incline. Calculate the net force on the mass. 10.5° What is the acceleration of the mass? 3.6 Students analyze motion of objects on inclined planes. 3.5 Now that we can find the net force of the object on the inclined plane, we can use our old equations to predict its motion IQ 11.14 Calculate the net force acting on the 3.5kg mass below. µ_k is 0.12. 52° 3.5kg If the mass is 0.38m up the ramp, how much time does it take for it to reach the bottom? 0.38m We can find acceleration by finding the net force. Once we have that, we use our old equations to find either time or velocity. A 1.2kg mass sits, initially at rest, 2.5m up a frictionless ramp with a 15° incline. What will be the velocity of the mass at the bottom of the ramp? 15° 1.2kg 2.5m Start by finding F_net Use F_net to find a Use velocity equation to find v ES 11.14 For the example problem to the left, how much time will it take for the block to reach the bottom of the ramp? 3.6 IQ 11.15 24° A 2.5 kg mass slides down a ramp with a 24° incline. If the ramp is 15m long, and µk = 0.15, calculate the time it takes for the mass to slide down the ramp. IQ 11.16 Below is a block being suspended from a rope while sitting on a frictionless ramp. Find a function for the tension on the rope in terms of mass of the block and theta Answer Quiz
When you are done you will turn it in. We will grade it tomorrow.

HW:
Read pgs. 147-149
copy the problem-solving strategies on pg. 149 in your own words.
try problem 1 on page 150

MQ like problem 1 tomorrow Open Note other practice problems:
pg. 165: 51, 52 MQ 11.17 65m A car in an action movie rides off a 65m high cliff at 14m/s. How far from the base of the cliff does the car land? ??? 14m/s 3.7 Students calculate changes in position seperately
in the horizontal and vertical directions.
3.8 Students calculate trajectories for projectiles and
can calculate maximum height, range, hangtime,
and a vertical position given its corresponding
horizontal position. Quiz 3 3.7 Students calculate changes in position seperately in the
horizontal and vertical directions. Just as with forces, we can split velocity apart in to horizontal and vertical Example: 22m/s 35° Draw a free body diagram for a baseball after it is hit. Ignore drag. What is the horizontal acceleration of this baseball?
What is the vertical acceleration of this baseball? Also, just like forces, we can calculate horizontal and vertical changes seperately.


¡Paramétrico de Ecuación! What will be the baseballs horizontal change in position over 1.0 seconds?
What will be the baseballs vertical change in position over 1.0 seconds? 3.8 Students calculate trajectories for projectiles and can
calculate maximum height, range, hangtime, and a
vertical position given its corresponding horizontal
position. Steps to solve these problems #1 What am I asking for?
max height
range
hangtime
corresponding position what is d when v =0? y y What is d when d =0? x What is t when d =0? y What is d when d = (some specific number) ? y x #2 What information do I have?
Look for v_0 and d_0 (those are very important)
Use what you decided in step #1
Split everything into horizontal and vertical #3 Solve for t
In every problem, you will have to solve for t.
It is the only thing that links the horizontal and vertical. #4 Find what you are looking for
look back at step #1
You will have to use t! Example:
An egg is thrown off a 25m high platform at 10.5m/s, 35° above horizontal. What will be its maximum height above the ground? ??? What is its range? ES 11.17 41° 24 m/s 8.0m The car jumps off the ramp as shown above. Answer each question below:
When the car lands, what will be d_y?
Fill out the table below (as much as you need to)
Find the hang time for the car
Find where the car will land
At its maximum height, what will be v_y?
maximum height 7.5m/s 21° 32m What will be the range of this water baloon? Put the steps in order.
Split all values into horizontal and vertical componants.
Use the time to find d_x.
Use the quadratic formula to find the time when d_y=0.
Find all position, velocity, and acceleration values. IQ 11.28
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