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Torque
It is easier to open a door when a force is applied at the
knob as opposed to a position closer to the hinges.
The farther away the force, the more torque there is.
Torque – the force(s) that cause an object to rotate
Torque is the product of the force and the perpendicular
distance from the axis of location
τ = rFsinӨ
τ = torque (mN)
r =moment arm (m)
F = force (N)
The force and the moment arm
should always be perpendicular to
each other.
Method A
The force is extended to meet the
moment arm which is drawn
perpendicular to force through the
hinge.
τ = rsinθF
Method B
The force is resolved into
components and the perpendicular
component is used with the
original moment arm.
τ = rFsinθ
Either method for calculating torque is acceptable
Two forces (Fa = 30 N and Fb = 20 N) are applied to a
meter stick which rotates around its left end. Which force
exerts the greater torque? What would be the net torque
if they were applied together?
Calculate the net torque acting on two wheels that are
attached. Ra is 30 cm and rb is 50 cm. Each force applied
is 50 N
Two children are playing on a seesaw which has a mass of 2.0 kg..
If child A (m = 30 kg) sits 2.5 m away from the center, at what
distance must child B (m = 25 kg) sit in order to balance the
seesaw?
What is the normal force acting on the seesaw?
Rotational Equilibrium
Equilibrium problems are ones where the objects in
question are stationary.
Is there a net force acting on
the traffic light?
No…the light is stationary.
ΣFx = 0 and ΣFy = 0
If the two forces are
equal…
Is there a net force?
No
Will it move?
Yes, because there
is a net torque.
A 1500-kg beam, 20.0 m long,
supports a 15,000-kg printing
press 5.0 m from the right
support column. Calculate
the force on each support
column.
When doing equilibrium problems involving rotation….
1. Draw a FBD showing all forces.
2. Choose a convenient coordinate system
3. Write the net force equations for the forces (x and y).
They should all be equal to zero.
4. Write the net torque equation. It should also be equal
to zero. Choosing a strategic axis of rotation eliminates
unknown forces.
5. Solve for the unknowns.
A 1500-kg beam, 20.0 m long,
supports a 15,000-kg printing
press 5.0 m from the right
support column. Calculate
the force on each support
column.
A uniform beam, 2.20-m
long with a mass of 25.0
kg (m), is mounted by a
hinge on a wall. The
beam is held horizontally
by a cable that makes an
angle of 30 degrees as
shown. The beam
supports a sign with a
mass of 28.0 kg (M).
Determine the force
exerted by the hinge on
the beam and the tension
force in the cable.
Circular Motion
This object is moving in a circular path at a constant speed.
The string is constantly
exerting a force on the
object
At any point in time, there
is a net force on the object
(no opposite forces)
According to Newton’s
First Law, the object must
be accelerating.
How? It is moving at a
constant speed.
The direction is changing, so the velocity is changing
Planets move
around the Sun in
a circular path
because of gravity.
The planets have a
radial or tangential
speed around the
orbit.
The force that is
causing the motion
is directed toward
the center.
aR = v2/r
aR = radial (centripetal) acceleration (m/s2)
v = radial speed (m/s)
r = radius of circle (m)
Centripetal forces cause centripetal accelerations.
Both are vector quantities pointed toward the
center of the circle.
A centripetal force is not a new type of force. Many
of the forces we have discussed are centripetal
forces when they cause objects to move in circles.
Σ FR = maR = mv2/r
Σ FR = net centripetal force (N)
m = mass (kg)
aR = centripetal acceleration (m/s2)
A 150-g ball at the end of a string is revolving
uniformly in a horizontal circle of radius 0.600 m.
The ball makes 2.00 revolutions per second. What
is the centripetal acceleration?
A 0.150-kg ball on the end of a 1.10-m cord is
swung in a vertical circle. What is the minimum
speed the ball must have at the top of its arc so
that the ball continues moving in a circle?
What would be the tension of the cord at the
bottom if the ball is traveling twice the speed of
the 1st part?
A 1000-kg car rounds a curve on a flat road of radius 50 m at
a speed of 14 m/s. Assuming the pavement is dry (μs =
0.60), will the car follow the curve or will it skid?
What if the pavement was covered
with ice (μs = 0.25)?
Banked Turns
Banked turns assist in helping cars navigate turns when there
is not sufficient friction for excessive speeds. How much of an
angle is needed so that friction is unnecessary?
How much of a banked turn
would be needed for a 1000-kg
car to steer through a 50-m
radius turn at a speed of 14
m/s?
The mass is irrelevant. Why?
Newton’s Law of Gravitation
Newton wondered why if an apple falls out of a tree toward
the Earth why the moon doesn’t fall toward the Earth also.
.
The moon actually falls, but because it is not stationary (it
is moving with a straight constant velocity), it follows a
curved path around the Earth.
Newton theorized that the
Moon did not get attracted
with the same force as the
apple nor did it fall with the
same gravitational
acceleration. Why not?
1. The moon was much
farther away from the Earth
than an apple on the surface.
2. The moon was much larger
than the apple.
Fg = Gm1m2/r2
G = gravitational constant = 6.67 x 10-11 Nm2/kg2
m1 = mass #1 (kg)
m2 = mass #2 (kg)
r = distance between centers of mass (m)
Newton’s Law of Universal Gravitation applies between
any two objects that have mass regardless of size.
A 50-kg woman sits 0.5 meters away from a 75-kg man.
What is the gravitational force between the two people?
To find out the acceleration of something falling on the
surface of the Earth, Newton’s 2nd Law can be applied.
Fg = Gm1m2/r2
mg = GmEm/rE2
g = GmE/rE2
g = gravitational acceleration (m/s2)
mE = mass of Earth (kg)
r = radius of Earth (m)
How is a ball’s gravitational acceleration on Mt. Everest
(altitude = 8850 m) different from its gravitational
acceleration at sea level?
Satellite Motion
Satellites are
objects that orbit
the Earth. They
are given a
radial speed
that counteracts
the effects of
gravity so that
they maintain a
circular path.
People that are in orbit
experience “apparent
weightlessness”.
It is not “actual weightlessness”
because the satellite and
occupants are still under the
effects of gravity.
There is NO normal force
pushing back on them giving
them a sense of having weight.
It is like the apparent
weightlessness felt by
occupants of an elevator
in freefall.
At what speed would a space station travel to maintain a circular
orbit around the Earth at an altitude of 4500 km? What would be
the gravitational acceleration of the space station?
Earth
Why do the occupants
experience weightlessness if
they are under the effects of
gravity?
Astronauts train in a similar
weightless environment in a
“zero g” airplane.
Plane makes large parabolic
turns with 30-s intervals of
weightlessness.
What do they experience at the bottom of the curve?
Simulated Gravity in Space
Physicists have theorized that floating (NOT orbiting) space
stations can be made to simulate gravity for the occupants
inside
These spacecraft are not significantly influenced by any type
of gravitational force – a simulated gravity is created by
rotating the space station around their axis
At what rotational speed (RPM) must a cylindrical
spaceship rotate if occupants are to experience simulated
gravitational acceleration like that of Earth? Assume the
spaceship’s radius is 500 m.