Forces

Documents you may need:

• Force vs Acceleration Lab (pdf or google doc)

• Mass vs Acceleration Lab (pdf or google doc)

• Mini Elevator Lab (pdf or google doc)

• Force on an oscillating spring lab (pdf or google doc)

• Friction Lab (pdf or google doc)

• Tension Lab (pdf or google doc)

• Extra Practice on Newton's Third Law: Forces Pairs Practice (pdf or google doc)

• "Elevator" Study Guide available as a pdf or google doc

• "Pulley" Study Guide available without friction (pdf) and with friction (pdf)

• Ch. 4 Study Guide & Puzzle (pdf or google doc) I strongly suggest the pdf!

• Net Force Practice worksheet (pdf or google doc although pictures may not come through)

• Forces Study Guide (pdf)

• Kinematics Equations + Forces Partner Practice (done in class while I was gone) (pdf or google doc) and the answers to it are here as a talking tutorial.

Homework Hints: Jump to today

November 3rd, 2014: Forces Test Study Materials

The Test Practice sheet (pdf or google doc) and the rubric (don't download it til you've tried the whole thing!) is available here.

Additional Problems you can do for Practice (approximate answers in parentheses):

Extra Problems you might want to try:

• pg. 142 #90 (0.255), 93 (370)

• pg. 130 #24 (5.5), 25 (0.5), 29 (1.3)

Types of forces problems:

- vertical net force (elevator & lifting)

- horizontal net force including Friction problems

- horizontal net force problems with more than one force (like tug of war)

- pulley problems (frictionless with/without an applied force)

Terms to know:

- Equilibrium

- Force

- Mass

- real Weight (Force of Gravity)

- Normal Force (apparent weight)

- Force of Tension

- Applied Force

- Force of Friction

- coefficient of friction

- net force aka "sum of the forces"

- Newton's First Law

- Newton's Second Law

- Newton's Third Lab

- UNITS! N, kg, m/s^2, etc

- constant velocity

Kinematics Equations + Forces Partner Practice

Kinematics Equations + Forces Partner Practice (done in class while Brix was gone) (pdf or google doc) and the answers to it are here as a talking tutorial. You'll have to download the pdf in order for it to talk to you.

pg. 130 #26 and pg. 142 #94

26. When Ke Min's brakes lock up the only force acting on his car is the force of friction. There are several steps to this problem:

(1) Solve for the Force of Friction with the given coefficient of friction and the Normal Force that you can find using the mass provided. (<10,000)

(2) Since that Force of Friction is the only force, it is also the FNet acting on the car. Given the mass and Force of Friction you can solve for the acceleration the car experiences. (>4)

(3) With that acceleration and the other kinematics information provided, use the green equation to solve for the displacement the car travels before it comes to a stop. If it is greater than 60 m Ke Min would hit the tree branch first.

94. This is a similar question to #26 BUT in this one you are looking for the coefficient of friction. Use the Kinematics information provided to solve for the acceleration. Since that Force of Friction is the only force, it is also the FNet acting on the car. Given the mass and acceleration you can solve for the Force of Friction the car experiences. The Normal Force that you can find using the mass provided and use that to find the coefficient of friction. (<0.5)

pg. 130 #22, 23 and pg. 142 #91 & 92

22. Given the block's mass you can solve for its Force of Gravity and since its not accelerating vertically you know the Normal Force is equal but opposite to the Force of Gravity. There is only one force acting on this block, the Force of Friction. That force is equal to the block's mass multiplied by its acceleration (and equal to the net force). The only thing missing would be the coefficient of Friction. (<0.2)

23. Given the book case's mass you can solve for its Force of Gravity and since its not accelerating vertically you know the Normal Force is equal but opposite to the Force of Gravity. Your applied force is being hindered by the Force of Friction; sum those two forces to get the net force which is equal to the bookcase's mass multiplied by the acceleration. The only thing missing would be the coefficient of Friction. (<0.2)

91. Same set-up as #23 except you are looking for the acceleration this time. Your equation should look the same. (>1)

92. a. Sum your forces and you'll add the applied force to your unknown force of friction and setting it equal to ma. Solve for the Force of Friction. (10) b. With that force of friction, use the equation and the normal force (which you can find with the mass) and solve for the coefficient of friction. (0.2)

pg. 128 #17-20

17. Remember that in order to pull the sled at a constant velocity the acceleration is 0 m/s^2. Given the weight of the sled you can determine the Normal Force applied to i t. Sum the two forces (Ff and FA) to determine the magnitude of the Force of Friction. Since you know the Normal force you can solve for the coefficient of friction. (<0.7)

18. Given the mass of the sofa find the Force of Gravity of the sofa. Use this to determine the Normal Force. Just like #17, sum your forces and find the coefficient of friction. (<0.1)

19. Given the weight (Fg) of the books, you can solve for the Normal Force since you know the box of books is not accelerating vertically. Given the coefficient of static friction, solve for the Force of Friction. (>70)

20. Since there is now someone sitting on the sled the Normal Force has increased, find the new Normal Force. Then sum the two forces (FA and Ff); remember the Net Force is 0 N. Solve for the size of the applied force. (<90)

pg. 130 #27, 28, 31

#27, 31 are conceptual

28. Given the boy's mass, solve for his Force of Gravity. This is equal but opposite to his Normal Force because he's not accelerating vertically. Use this Normal Force and the coefficient of friction provided to solve for the Force of Friction. (<40)

rest of Net Force Practice worksheet #6-9

6. Parts (a) and (b) are important for determining the equation for part (c) correctly. There are only two forces acting on the picnic basket. Solve for the force required to lift the basket up (calling it FA or FT is fine). (>200)

7. Parts (a), (b) and (c) are important for determining the equation for part (d) correctly. The force of gravity on the pumpkin is the only force in the direction of motion because the two forces of tension are equal but in opposite directions. The pumpkin's weight is pulling both the pumpkin and the squash down. Since they are going down the acceleration should be negative. Since the pumpkin is tied to the squash it is not in free fall and its acceleration should be less than the acceleration due to gravity (-9.8m/s^2). (d) (<5)

8. Parts (a), (b) and (c) are important for determining the equation for part (d) correctly. In part (c) the applied force of the stunt man (on top of the cliff) is working against the force of gravity. Those two forces sum together to make the net force. In order for this to be a "rescue" the acceleration would have to be up the cliff (positive) and so would the net force (if the force of gravity on the daredevil were winning, it wouldn't be a rescue). (d) (>0 but not by much).

9. Parts (a), (b) and (c) are important for determining the equation for part (d) correctly. Looking at Atwood's Machine, you can imagine straightening out the string with the same forces still being applied. Refer to the picture below. Note that one of those forces will have to be considered negative and one will have to be considered positive because they are pulling in opposite directions. Comparing the two masses you should be able to guess which direction will "win" the tug-of-war and which way it will accelerate. (<2)Net Force Practice worksheet #1-5 1. There are two forces acting on a stone. When its "weight" is given as 5.4 N that's the Force of Gravity. Use Fg=mg with this given Force of Gravity to find the mass. (<1) There must be a net force since its accelerating up. The applied force that is lifting it up must be larger than the Force of Gravity. Use Fnet = FA + Fg = ma. (<10) 2. Same set up as #1 but you are given the mass (use that to find the Fg) and the applied force. You're looking for the acceleration this time. (<1) 3. Same set up as #1. (>3000) 4. Given the boy's mass solve for his Force of Gravity (Fg=mg). You can sum the two forces involved and solve for the acceleration using Fnet = FA + Fg = ma. (<5)

5. Given the man's Force of Gravity, solve for his mass using Fg=mg. Then sum the two forces and solve for the acceleration. (<5)

October 21st, 2014: pg. 105 #34-38, 79

#34 & 35 are conceptual - If you're having trouble even after we talked about it try this:

36. Make sure you solve for both forces of tension, the string on the top and the bottom. The bottom string is working only against the force of gravity of the bottom block (5kg). The top string is working against the force of gravity of the bottom and top block (10 kg).

37. This time you are given the force of tension in the top block (63N) and the size of the bottom block. Since you know that the top force of tension is working against both force of gravity, you can subtract the force of gravity for the bottom block (since you know its mass). The difference is the force of gravity on just the top block. Use Fg=mg to solve for that unknown block. (>3)

38. Remember that the Normal Force is the force that a surface has to push back up against the Force of Gravity.

79. Similar to the "hanging" problems you did above but this time there is an acceleration and therefore there is a net force. Add the two forces involved and set them equal to "ma." Solve for the missing force.

pg. 97 #15-18 and pg. 101 #21, 23, 25