Download Class Notes

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Specific impulse wikipedia , lookup

Fictitious force wikipedia , lookup

Force wikipedia , lookup

Vibration wikipedia , lookup

Newton's laws of motion wikipedia , lookup

Classical central-force problem wikipedia , lookup

Kinematics wikipedia , lookup

Jerk (physics) wikipedia , lookup

Equations of motion wikipedia , lookup

Work (physics) wikipedia , lookup

Inertia wikipedia , lookup

Modified Newtonian dynamics wikipedia , lookup

Rigid body dynamics wikipedia , lookup

Relativistic mechanics wikipedia , lookup

Seismometer wikipedia , lookup

Centripetal force wikipedia , lookup

Mass versus weight wikipedia , lookup

Center of mass wikipedia , lookup

Transcript
Type: Double
Date: ______________
Objective: Statics and Dynamics
Homework: Assignment (1-20)
Do PROBS Ch. 4 #’s (82) Ch. 4 +
Do AP 1986 #1, 1988 #1 (handouts)
AP Physics “B”
Mr. Mirro
Date: ________
Statics and Dynamics I
Two masses suspended over a pulley by a cable, as shown is sometimes referred to generically as an
Atwood’s machine.
Consider the real-life application of an “elevator” (m1) and its “counterweight” (m2).
To minimize the work done by the motor to raise and lower the elevator safely, m1 and m2 are
similar in mass.
We leave the motor out of the system for this calculation, and assume the cable’s mass is negligible
and the pulley is frictionless and massless, which assures that the “tension” in the cord has the
same magnitude on both sides of the pulley.
Let the mass of the counterweight be m2 = 1000 kg. Assume the mass of the empty elevator is 850 kg, and
its mass when carrying four passengers is m1 = 1150 kg.
For the latter case (m1 = 1150 kg), we can determine:
a. the acceleration (a) of the elevator
motion
b. the tension in the cable (T)
adown = negative
FIRST – The Analysis of the Elevator (m1):
∑ FE = mE adown
aup = positive
mE = m1
+T - mE g = mE (-a)
SECOND – The Analysis of the Counterweight (m2):
∑ FCW = mCW aup
mCW = m2
+T - mCW g = mCW (+a)
Finally, solving the system of equation by either addition or substitution we can obtain an expression
for the acceleration (a).
Method #1: Solving systems of equations by using the addition method.
+T - mE g = mE (-a)
Eq. 1
+T - mCW g = mCW (+a)
Eq. 2
Multiply Eq. 1 through by (-1) to switch the signs of each term to create
a new version of Eq 1 called Eq. 3.
-T + mE g = mE (a)
ADD
Eq. 3
Now we can add Eq. 2 and the new version of Eq. 1, which now is Eq. 3 together to
eliminate the tension (T) and then isolate the acceleration (a).
+T - mCW g = mCW (a)
Eq. 2
-T + mE g = mE (a)
Eq. 3
+ mE g - mCW g = mCW (a) + mE (a)
mE g - mCW g = a (mCW + mE )
a = mE g - mCW g
mCW + mE
Method #2: Solving systems of equations by using the substitution method.
Since, +T - mE g = mE (-a) then:
T1 = mE g - mE a
Since, +T - mCW g = mCW (+a) then:
T2 = mCW a + mCW g
Let
T1 = T2
mE g - mE a = mCW a + mCW g
By substitution of T1 and T2
mE g - mCW g = mCW a + mE a
Same as previous !
mE g - mCW g = a (mCW + mE)
Same as previous !
a = mE g - mCW g
mCW + mE
Same as previous !
AP Physics “B”
Mr. Mirro
Date: ________
Statics and Dynamics I
Ex 1: As shown in the diagram, two guiderails for an elevator
each exert a constant friction force of 95.0 Newtons on the
elevator car when it is moving upward with an acceleration
of 1.0 m/s2 . The pulley has negligible mass and friction;
assume g = 10.0 m/s2 .
a) What is the direction of the net force on the elevator car?
500 kg
Elevator
Guide Rails
b) On the diagram below, sketch and clearly label all the
forces acting on the elevator car.
M
Counterweight
c) Calculate the tension in the cable when it is accelerating upward as described.
d) Calculate the mass M the counterweight must have in order to raise the elevator car with the acceleration as
described.
Ex 2: A helicopter holding a 50-kilogram package suspended from a rope 10.0 meters long accelerates
upward at a rate of 6 m/s2. Neglect air resistance on the package.
a.
On the diagram below, draw and label all of the forces acting on the package.
b. Determine the tension in the rope.
c. When the upward velocity of the helicopter is 20 meters per second, the rope is cut and the helicopter
continues to accelerate upward at 6 m/s2. Determine the distance between the helicopter and the
package 2.0 seconds after the rope is cut.
AP Physics “B”
Mr. Mirro
Date: ________
Statics and Dynamics II
Ex: A crate of dilled pickles with mass m1 = 14 kg moves along a plane that makes an angle
of θ = 30° with the horizontal. That crate is connected to a crate of pickled dills with mass
m2 = 14 kg by a taut, massless cord that runs around a frictionless, massless pulley. The
hanging crate of dills descends with constant velocity. [Halliday6.4]
a. What are the magnitude and direction of the frictional
force exerted on m1 by the plane ?
m2
m1
b. Compute the value for the coefficient of kinetic friction between the crate and the plane.