Torque | MCAT 2015
Torque – a “twisting force”
• A net force causes an object to
travel along a curve
Converting the F Vector into Components
Converting the force vector into components, we
are left with 2 components that are either
perpendicular or parallel to the position vector.
Boy on a Seesaw
The boy’s downwards gravitational force induces
a torque, which is responsible for the impending
counterclockwise motion of the seesaw.
•
τ = Frsinθ
τ = torque
F = force
r = position vector
θ = angle between those vectors
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θ is now 90°, and sin90° equals 1
So, we can simplify our equation:
o τ = Fl
τ = torque
F = force
l = lever arm
The force is only the force
perpendicular to the position
o The horizontal force
component doesn’t
contribute to torque
The lever arm is just the length
of the position vector
Calculating the Torque
• Suppose:
o The boy’s gravitational
force is 150 N
o The lever arm is 2 meters
o The seesaw’s tilt is 30°
Identifying the Variables Required to
Calculate Torque
•
As we’ve seen in so many
situations before, calculations are
greatly simplified when we can
make our vectors perpendicular
o By converting the force
vector into components,
we can make F and r
perpendicular
Determining the Angle Between the Force
Vector and its Components
Evidently, tilting the seesaw 30° will produce a
30° angle (highlighted) between the downwards
force vector (faded blue) and its component
vector (opaque blue)
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© 2016 J Co Review, Inc., Accessed by Guest on 10-01-2016
Torque | MCAT 2015
•
Knowing the 30° angle, we could
use trigonometry to find our
component vectors
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Finding Component Vectors
Using the 30° angle, you can use SOH CAH
TOA to find component vectors of 130N & 75N.
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With a lever arm (2 m) and a
perpendicular force component
(130 N), we can solve for torque
τ = (130 N)(2 m) = 260 Nm in
the counterclockwise direction
In this case, there is torque in
only one direction
•
•
But, we now have to find 3
torques
For the 8 kg weight:
o F = mg = 80N
o l=3m
o τ = 240 Nm
For the 6 kg weight:
o F = mg = 60N
o l=1m
o τ = 60 Nm
For the 10 kg weight:
o F = mg = 100N
o l=2m
o τ = 200 Nm
The 8 kg and 6 kg weights
produce torques
counterclockwise, so we can sum
their torques to 300 Nm
The 10 kg weight produces a
clockwise torque
Torque in Both Directions
• A plank hanging from the ceiling
suspends three weights
o They are spaced from the
center as shown
Final Torques
Adding the two counterclockwise torques, the
final counterclockwise torque is 300 Nm. The
clockwise torque is 200 Nm.
•
Plank Suspending Three Different Weights
•
Our three forces are already
perpendicular to the lever arm
o We don’t need to find
component vectors
•
Since the counterclockwise
torque outweighs the clockwise
torque, the plank will rotate
counterclockwise
Why didn’t we calculate the
torque of the gravitational force
on the plank itself?
o FG acts on the plank’s
center- at point of rotation
o Lever arm=0, and so τ = 0
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© 2016 J Co Review, Inc., Accessed by Guest on 10-01-2016