TORQUE

1
TORQUE This topic relates to Rigid Body
Systems. Rigid Body A system that has
a size that must be considered, and a shape that
does not change under the influence of forces. A
rigid body is a kind of idealized solid in which
the distortion that real bodies undergo
(stretching, bending, twisting) are imagined
not to occur. Rigid bodies undergo
translational motion. (all points in the body
move along parallel paths) Rigid bodies undergo
rotational motion. (when the body spins about a
balance point.) Since rigid bodies can rotate
as well as move translationally, they are in
equilibrium when they are at rest or in uniform
rotational or translational motion. Levers are
examples of rigid body systems in which the pivot
point is called the fulcrum.
2
Torque The torque generated by a force about a
given point is, the force applied multiplied by
the distance from the pivot along a line
perpendicular to the direction of the
force. Torque is the product of the force and
the effective length of the torque arm.
(distance between the pivot and applies force)
? Fd?
3

• Equilibrium of a rigid body is satisfied when the
  following conditions are met
• The resultant of all translational forces
• acting on a body equals zero.
• F? F?
• 2. The resultant of all the torques (about a
  pivot) due to forces applied equals zero.
• ? cw ? ?ccw ?

4
Examples
Translational
F4
F5
F1
F2
F3
F1 F2 F3 F4 F5
5
Samples
Rotational
? cw ? ?ccw ?
d2
d1
?
d3
F1
F2
F3
F2d2 F1d1 F3d3
6
Examples
Rotational
? cw ? ?ccw ?
d2
?
L2
d1
F2
?
?
L1
F2d2 F1d1
F1
F2(L2cos?) F1( L1cos?)
7
Examples
Rotational
F1
? cw ? ?ccw ?
d
?
?
?
F?
d?
? F(dsin?)
? Fd?
OR
? (Fsin?)d
? F?d
8
Sample Problem
A 100 m uniform bridge weighing 4.0 x105 N. is
supported on either end by two cement abutments.
A car weighing 2.0 x 103 N. is on the bridge 30
m from the right end. Calculate the force on
each abutment (Fa Fb).
dtotal 100 m dcar 30 m from
right. Fbridge 4.0 x105 N. Fcar 2.0 x
103 N. Fa ? Fb ? Diagram
Uniform means, center of gravity is in the
middle of the bridge
30 m
Fa
Fb
100 m
2.0 x 103 N.
4.0 x 105 N.

• The bridge is not moving so it is in Equlibrium !
• Look at Translational Equlibrium
• Fa Fb 4.0 x 105 N. 2.0 x 103 N.

9
Sample Problem
dtotal 100 m dcar 30 m from
right. Fbridge 4.0 x105 N. Fa ? Fcar
2.0 x 103 N. Fb ?
30 m
Fa
Fb
100 m
2.0 x 103 N.
4.0 x 105 N.

• The bridge is not moving so it is in Equlibrium !
• Look at Translational Equlibrium
• Fa Fb 4.0 x 105 N. 2.0 x 103 N.
• So
• Fb 4.0 x 105 N. - Fa
• Now use Rotational Equilibrium to find Fa.
• ? cw ? ?ccw ?

10
Sample Problem
dtotal 100 m dcar 30 m from
right. Fbridge 4.0 x105 N. Fa ? Fcar
2.0 x 103 N. Fb ?
30 m
Fb
Fa
100 m
2.0 x 103 N.
4.0 x 105 N.

• Fb 4.0 x 105 N. - Fa
• Now use Rotational Equilibrium to find Fa.
• ? Fd? and ? cw ? ?ccw ?
• Placing the pivot at either end ie.
  (Right)
• The equation above reads
• Fa x 100m (4.0 x 105 N. x 50m) (2.0 x 103 N.
  x 30m)
• Placing the pivot at Fb causes the torque from
  Fb 0

11
Sample Problem
dtotal 100 m dcar 30 m from
right. Fbridge 4.0 x105 N. Fa ? Fcar
2.0 x 103 N. Fb ?
30 m
Fb
Fa
100 m
2.0 x 103 N.
4.0 x 105 N.

• Fb 4.0 x 105 N. - Fa
• Fa x 100m (4.0 x 105 N. x 50m) (2.0 x 103 N.
  x 30m)
• Thus
• Fa (4.0 x 105 N. x 50m) (2.0 x 103 N. x 30m)
• 100m
• Fa (2.0 x 107 N.m) (6.0 x 104 N.m)
• 100m
• Fa 2.006 x 105 N
• Substitute back into the first equation and solve
  for Fb

12
Sample Problem
dtotal 100 m dcar 30 m from
right. Fbridge 4.0 x105 N. Fa ? Fcar
2.0 x 103 N. Fb ?
30 m
Fb
Fa
100 m
2.0 x 103 N.
4.0 x 105 N.

• Fb 4.0 x 105 N. - Fa
• Fa 2.006 x 105 N
• Fb 4.0 x 105 N. - 2.006 x 105 N
• Fb 1.99 x 105 N