Know Definitions of Key
   Terms & Symbols
Focus during the entire Power Point activity.
   Solidify your studying skills during this
   class period.
Perform your work in your science journal so
   you have created a study guide for the test.
Call me over if you are having difficulty
   getting started.
If your answer is confirmed as correct, become
   a student/teacher and help someone in class
   who does not understand the method used
   to solve the problem.
Batfink, who has a mass of 75 kg is
placed in a 25 kg stationary barrel.
What is the Fg on Batfink and the
barrel?
SOLUTION:
Force of gravity on Batfink and the barrel.


mbf = 75 kg        Fg
mb = 25 kg
g = -9.8 m/s2

                   Fg= -980 N
Batfink and the barrel are raised at
1.1 m/s2. What is the force of
support acting on Batfink and the
Barrel?
SOLUTION:
Force of support on Batfink and the barrel.


mbf = 75 kg        Fs
mb = 25 kg
g = -9.8 m/s2
a = 1.1 m/s2
                   Fs= 1090 N
Suddenly, Batfink and the
barrel are lowered at 1.1
m/s . What is the force of
    2


support acting on Batfink
and the Barrel?
SOLUTION:
Force of support on Batfink and the barrel.


mbf = 75 kg        Fs
mb = 25 kg
g = -9.8 m/s2
a = -1.1 m/s2
                   Fs= 870 N
Hugo Ago-go pushes the barrel with
Batfink in it towards the end of the
cliff with a 100 N force over a
distance of 10 m before the barrel
leaves the cliff. The force of friction
is 8 N. Draw a force diagram of the
situation.
y

                           Fs 980 N


Ff - 8 N              Fa 100 N   x


               Fg -980 N
Calculate the acceleration
of the barrel in the x
axis.
SOLUTION:
Acceleration of Batfink and the barrel.


mbf = 75 kg        a
mb = 25 kg
g = -9.8 m/s2
Fa = 100 N
Ff = 8 N
Xi = 0 m
X = 10 m          a = 0.92 m/s2
What is the Vf of the
barrel just before it falls
off the cliff?
SOLUTION:
Final velocity of Batfink and the barrel.


Vi = 0 m/s         Vf
a = 0.92 m/s2     tf
Xi = 0 m
Xf = 10 m
Ti = 0 s
                 V = 4.29 m/s
The Incredible Hulk is hanging motionless off
the ground by chains attached separately to
his wrists from two different walls. The Hulk
has a mass of                  355 kg. The
chain on his                     right wrist
(T1) forms an                     angle of 26˚
relative to the                floor, and the
chain from his             left wrist (T 2)
forms an angle              of 32˚ relative to
the floor. T2 has             2500 N acting
on it. Draw a                 force diagram
of the situation.
y

                    T1
T2 = 2500 N


              26°        32°



Fg -3479 N
Determine the tension in T2X.
SOLUTION:
Tension in chain #1. First find T2x.


 T2 = 2500 N        T2x
 θ = 32°
m = 355 kg
g = -9.8 m/s2
Fg = -3479 N
                 T2x = 2120 N
SOLUTION:
Tension in chain #1.


 T2 = 2500 N           T1
 θ = 32°
m = 355 kg
g = -9.8 m/s2
Fg = -3479 N
T2x = 2120 N     T1 = 2358.2 N
7 2012 ppt force review

7 2012 ppt force review

  • 2.
    Know Definitions ofKey Terms & Symbols
  • 3.
    Focus during theentire Power Point activity. Solidify your studying skills during this class period. Perform your work in your science journal so you have created a study guide for the test. Call me over if you are having difficulty getting started. If your answer is confirmed as correct, become a student/teacher and help someone in class who does not understand the method used to solve the problem.
  • 4.
    Batfink, who hasa mass of 75 kg is placed in a 25 kg stationary barrel. What is the Fg on Batfink and the barrel?
  • 5.
    SOLUTION: Force of gravityon Batfink and the barrel. mbf = 75 kg Fg mb = 25 kg g = -9.8 m/s2 Fg= -980 N
  • 6.
    Batfink and thebarrel are raised at 1.1 m/s2. What is the force of support acting on Batfink and the Barrel?
  • 7.
    SOLUTION: Force of supporton Batfink and the barrel. mbf = 75 kg Fs mb = 25 kg g = -9.8 m/s2 a = 1.1 m/s2 Fs= 1090 N
  • 8.
    Suddenly, Batfink andthe barrel are lowered at 1.1 m/s . What is the force of 2 support acting on Batfink and the Barrel?
  • 9.
    SOLUTION: Force of supporton Batfink and the barrel. mbf = 75 kg Fs mb = 25 kg g = -9.8 m/s2 a = -1.1 m/s2 Fs= 870 N
  • 10.
    Hugo Ago-go pushesthe barrel with Batfink in it towards the end of the cliff with a 100 N force over a distance of 10 m before the barrel leaves the cliff. The force of friction is 8 N. Draw a force diagram of the situation.
  • 11.
    y Fs 980 N Ff - 8 N Fa 100 N x Fg -980 N
  • 12.
    Calculate the acceleration ofthe barrel in the x axis.
  • 13.
    SOLUTION: Acceleration of Batfinkand the barrel. mbf = 75 kg a mb = 25 kg g = -9.8 m/s2 Fa = 100 N Ff = 8 N Xi = 0 m X = 10 m a = 0.92 m/s2
  • 14.
    What is theVf of the barrel just before it falls off the cliff?
  • 15.
    SOLUTION: Final velocity ofBatfink and the barrel. Vi = 0 m/s Vf a = 0.92 m/s2 tf Xi = 0 m Xf = 10 m Ti = 0 s V = 4.29 m/s
  • 16.
    The Incredible Hulkis hanging motionless off the ground by chains attached separately to his wrists from two different walls. The Hulk has a mass of 355 kg. The chain on his right wrist (T1) forms an angle of 26˚ relative to the floor, and the chain from his left wrist (T 2) forms an angle of 32˚ relative to the floor. T2 has 2500 N acting on it. Draw a force diagram of the situation.
  • 17.
    y T1 T2 = 2500 N 26° 32° Fg -3479 N
  • 18.
  • 19.
    SOLUTION: Tension in chain#1. First find T2x. T2 = 2500 N T2x θ = 32° m = 355 kg g = -9.8 m/s2 Fg = -3479 N T2x = 2120 N
  • 20.
    SOLUTION: Tension in chain#1. T2 = 2500 N T1 θ = 32° m = 355 kg g = -9.8 m/s2 Fg = -3479 N T2x = 2120 N T1 = 2358.2 N