Transcript Torque - rcasao

Torque
Physics
Montwood High School
R. Casao
Torque
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As we have seen over and over again, a force is
a push or a pull, and an unbalanced force
causes an object to move (to accelerate).
All the motion we have dealt with so far has
moved objects from one point in space to
another. This type of motion involving a
displacement is called translational motion.
We will now consider another type of motion,
rotational motion, where an object moves not
from one place to another but rather around
some axis of rotation.
Translation (linear) Motion
Translation + Rotation
Torque
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Forces are also necessary for
rotational motion.
The rotational effect of a force
depends upon its point of
application.
Torque: is the rotational analog of
a force and measures the tendency
of a force to rotate a body around
a pivot (fulcrum). Actual rotation
does not need to occur.
Torque
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A torque is not a force.
Pivot point: the point from which the
lengths of all torque arms, also called
moment arms, are measured.
Once a pivot point is chosen, a
perpendicular line is drawn to each of
the lines along which the force vectors
act on the object. Torque arm or
moment arm is the perpendicular
distance from the pivot point to the line
of action of the force.
Torque
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The line of action of the force passes
through the point at which the force is
applied and is parallel to the force.
Calculating a torque always requires that
you choose one point as your only pivot
and calculate all torques relative to it.
Torque, the rotational effect of a force is
the product of a force and the length of
its torque arm.
Torque
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Torque = Force applied · torque arm
Mathematically:
T  F  r
Unit: m·N or N·m
When forces are not applied parallel
(00) to or perpendicularly (900) to the
torque arm, the mathematical
expression for the torque is:
T  F  r  sinθ
Torque
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where  is the angle between the line of
action of the force and a line drawn from
the point of application of the force to
the pivot.
The torque arm is once again the
perpendicular distance from the pivot to
the line of action of the force.
If the distance from the point of
application of the force is r, then the
torque arm is one side of a right triangle
in which r is the hypotenuse.
Torque (simplest case)
“Lever arm” or torque arm is perpendicular to force.
  F l
Units: N·m
Torque
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The torque for a force applied at the
pivot point is always zero. The moment
arm for any force applied at the pivot
point is zero.
Direction of torque:
– To identify a torque as clockwise or
counterclockwise, imagine that the object is
free to rotate around a stationary pivot point.
– Imagine that the force producing the torque
is the only force acting on the object.
– The direction in which the bar would rotate is
the direction of the torque.
Torque
Torque
How do we quantify torque when force is not
perpendicular to the lever arm?
Define:
• Line of Action:
The line along the force
vector.
• Lever Arm:
The distance between
the line of action and
the axis of rotation
measured along a line
that is perpendicular
to both.
Torque
Zero torque if force is
directed toward
center of
rotation.
Mechanics of Rigid Body
EQUILIBRIUM CONCEPT
Conditions for equilibrium
equilibrium = no acceleration, i.e. a particle remains at rest, or if in
motion, moves with constant velocity.
EQUATIONS OF MOTION OF A RIGID BODY
 Fext  m aCM
 CM  I CM 
 O  IO
CONDITIONS FOR EQUILIBRIUM
 Fext  O
  0
In the case of a rotation about a fixed axis
and O is a point of this axis. In general, if O
is a point of a inertial reference system
1. The net external force acting on
the body must remain zero.
2. The net external torque about any
point must remain zero. Any point
can be considered because all of
them are in a inertial reference
system (non accelerated).
Rotational Equilibrium
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An object at equilibrium has no net
influences to cause it to move, either in
translation (linear motion) or rotation.
The basic conditions for equilibrium
are:
Rotational Equilibrium
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A torque that tends to rotate
something counterclockwise is
usually defined as positive; a
clockwise torque is defined as
negative.
Rotational equilibrium occurs when
the angular acceleration and the
sum of the torques is equal to zero.
– Dynamic rotational equilibrium:
angular velocity is constant; angular
acceleration is zero.
Rotational Equilibrium
– Static rotational equilibrium: angular
velocity and angular acceleration is
zero.
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Three factors determine how
effective a torque is in altering the
rotational state of an object:
– the magnitude of the force,
– the direction of the force, and
– the point of application of the force.
Problem-Solving Tips
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All torques must be calculated
about the same pivot.
When a problem contains two
unknowns, mathematically remove
one of the unknowns by selecting
it as the pivot. This causes the
torque due to the applied force at
that location to be zero.
Coupled Forces
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Couple: two forces of equal magnitude acting
in opposite directions in the same plane, but
not applied at the same point.
Although the forces are in opposite directions,
each tends to rotate the body in the same
direction about the pivot.
The torque of a couple is the product of one of
the forces and the perpendicular distance
between them.
A couple can only be balanced by a second
couple; the torques of the two couples must
have equal magnitudes but opposite
directions.
The Lunch Tray Example
Your fingers and thumb both exert a force on
the tray that acts to keep it in equilibrium.
Tray: m = 0.2 kg; food: m = 1.0 kg;
cup: m = 0.250 kg
What is F and T?
The Lunch Tray: free-body diagram
F2
Wt
F1
Wc
Wf
Mechanics of Rigid Body: Statics
• Determine B by solving the
equation for the sum of the
moments of all forces about A.
Note there will be no
contribution from the unknown
reactions at A.
A fixed crane has a mass of
1000 kg and is used to lift a
2400 kg crate. It is held in
place by a pin at A and a
rocker at B. The center of
gravity of the crane is located
at G.
Determine the components of
the reactions at A and B.
• Determine the reactions at A
by solving the equations for
the sum of all horizontal force
components and all vertical
force components.
• Check the values obtained for
the reactions by verifying that
the sum of the moments about
B of all forces is zero.
Free-body diagram
First step in the static equilibrium analysis
of a rigid body is identification of all
forces acting on the body with a free-body
diagram.
• Select the extent of the free-body and
detach it from the ground and all
other bodies.
• Indicate point of application,
magnitude, and direction of external
forces, including the rigid body
weight.
• Indicate point of application and
assumed direction of unknown
applied forces. These usually consist
of reactions through which the ground
and other bodies oppose the possible
motion of the rigid body.
• Include the dimensions necessary to
compute the torque arms of the
forces.
Mechanics of Rigid Body.
SOLUTION:
• Create a free-body diagram of the
joist. Note that the joist is a 3
force body acted upon by the
rope, its weight, and the reaction
at A.
A man raises a 10 kg joist, of
length 4 m, by pulling on a rope.
Find the tension in the rope and
the reaction at A.
• The three forces must be concurrent
for static equilibrium. Therefore, the
reaction R must pass through the
intersection of the lines of action of the
weight and rope forces. Determine the
direction of the reaction force R.
Equilibrium Example: Ladder
WF=875 N
WL=355 N
dL=8.0 m
dF=6.3 m
Find the forces that the
wall and the ground
exert on the ladder.
The Ladder
Free-body diagram:
Ladder: sum of torques is zero
The normal force
exerted by the ground
on the ladder and the
frictional force
exerted by the ground
do not apply torques
to the ladder.