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| Date post: | 17-Nov-2014 |
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Property of Mark Joseph Bantayan1900 Philippine Normal University, Taft Ave Manila
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TORQUE
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What is Torque?
Imagine a point W on the wrench on the image to the left (10.01). If you apply a force of Fa * on point Wa you feel a great resistance to your effort to remove the bolt.
As you move the point of contact to Wb you will notice that there is less resistance to the force Fb. But if you apply Fc at Wc, your hands only slides past the wrench handle.
The resistance you felt was equal to the force you applied but opposite in direction, that is according to Newton’s Action-Reaction pair. Considering that you applied the same magnitude of Force on all three points yet got different results implies the existence of another quantity that causes such variation.
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That quantity is Torque. Commonly symbolized by Г (the Greek letter Gamma). It is the product of the Force (F) and lever arm (R) with the angle from the axis of rotation (sin Ф). It is useful to think of it as a rotational analogue of Work since the lever arm time the angle is equal to the linear displacement (although this is misleading and is actually a misconception perpetuated by public schools). Although the dimensions are almost the same, we do not use Joules as a unit. We use Newton-meters instead (N*m).
Г = r * F ( the torque vector)
Г = FL = rF sinf = Ftanr (magnitude of torque)
→ → →
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torque (moment of a force or couple):
The product of a force and its perpendicular distance from a point about which it causes rotation or *torsion. The unit of torque is the Newton meter, a vector product, unlike the joule, also equal to a Newton meter, which is a scalar product. A turbine produces a torque on its central rotating shaft. ~ Oxford Dictionary of Science
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PIVOT POINT
LEVER ARM
POINT OF EXERTION
TO FIND TORQUE: 200 CM
5 N
Г = F L (FORCE)(LEVER ARM)L = RsinФ (LEVER ARM )Г= (5 N) ( 200CM × 1 M ) (sin180°) = (5 N) [(2 M) (0)] = 0
Г = 0 Nm
Evaluate: Since the application of force is parallel to the lever arm, the system does not rotate. There is no torque.Also sin180 is 0, so the product of force and lever arm is 0.
100 CM
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0.0085 KM
44 NГ = F RsinФ
= (44 N) (0.0085 KM × 1 M ) ( sin 90°) 0.001
KM = (44 N) (8.5 M) (1) Г = 374 Nm
Evaluation: The force applied is perpendicular to the lever arm, therefore there is a tendency to rotate, therefore torque is present in the system. By definition torque is the product of force and lever arm. Since the sin of 90 is 1, the definition holds true for the example. PREV NEXT
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∏ 6
RADS
6 M
50 N
∏
6rads × 360°
2∏ rads
Ф =
= 30°
1 M
R = 6 M – 1 M = 5M
L = 5 M sin 30° = 2.5 MГ = 125 N M
Evaluate:
The two angles are similar (recall transversals, they always come up in quizzes, there may even be some problems that require hardcore triangle geometry)
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Torque and Angular Acceleration
Imagine a rigid body that rotates about a fixed axis z. Let this body be composed of mass particles from m1 to mn . Each of the mass particles has a distance r1 to rn from the axis rotation.
The z component of thenet force that acts on each mass particle is Ftan .
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Z axis here
R1
distance 1
M1
point mass
By First Newton’s Law:
Ftan = matan
r [ Ftan = matan ] r Ftan r = mr α since [atan = αr] [Ftan r = Г] Г = mr α [I = mr ] Г = I α
2
2
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R = r1 + r2 + r3 + … + rl + …
rn-1 + rn = ∑ri
M = m1 +m2 + m3 +…+ mn-l
+ mn = ∑mi
Z axis here
R1
distance 1
Rl
distance l
Rm
distance m
M1
point mass
Ml
point mass
Mm
point mass
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The sum of the distances from the z-axis:
∑ri = r1 + r2 + r3 + … + rl + … rn-1 + rn
The sum of the point masses :
∑mi = m1 +m2 + m3 +…+ mn-1 + mn
The sum of the net forces on each point masses:
∑F = m1atan1 +m2 atan2+ m3atan3 +…+ mn-
latan(n-1) + mnatann
The sume of Inertia experienced by each point mass :
∑ I = m1 r1 +m2 r2+ m3 r3 +…+ mn-1 rn-1 + mn rn
∑atan = αr1 + α2 r2 + α 3 r3 +…+ α (n-1) rn-1 + α
rn
∑F × ∑ri = m1atan1 r1 + m2atan2r2 + … + atan(n-1)rn-1 +atannrn
the total torque on a rigid body: ∑Г = I1α1 + I2α 2 + I3α 3 + In-1αn-1 + Inαn
∑Г = ∑F × ∑ri = ∑ I ×
∑ α
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Torque is equal to the product of the rotational inertia and the radial acceleration.
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TORQUE IN EVERYDAY LIFE:
Torque converters multiply the force of the car engine.
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OIL DIGGERS
CRANES
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Varying Distance
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CHOOSE R:
FORCE APPLIED:50 NEWTONS
90 DEGREES
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0.5 METER
1.00METER
1.5 METER
Г = (50 N)(?? M)(sin90°)
Torque = ?? NM
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CHOOSE R:
FORCE APPLIED:50 NEWTONS
90 DEGREES
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0.5 METER
1.00METER
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Г = (50 N)(1.5 M)(sin90°)
Torque= 75 NM
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CHOOSE R:
FORCE APPLIED:50 NEWTONS
90 DEGREES
HOME
0.5 METER
1.00METER
1.5 METER
Г = (50 N)(1.00 M)(sin90°)
Torque = 50 NM
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CHOOSE R:
FORCE APPLIED:50 NEWTONS
90 DEGREES
HOME
0.5 METER
1.00METER
1.5 METER
Г = (50 N)(0.5 M)(sin90°)
Torque = 25 NM
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CHOOSE F:
1.00 M
85 DEGREES
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150 NEWTONS
100 NEWTONS
50 NEWTONS
Г = (?? N)(1.00 M)(sin85°)
Torque = ?? NM
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CHOOSE F:
1.00 M
85 DEGREES
HOMEГ = (150 N)(1.00 M)(sin85°)
Torque = 49.80NM
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100 NEWTONS
50 NEWTONS
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CHOOSE F:
1.00 M
85 DEGREES
HOMEГ = (100 N)(1.00 M)(sin85°)
Torque = 99.62 NM
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CHOOSE F:
1.00 M
85 DEGREES
HOMEГ = (150 N)(1.00M)(sin85°)
Torque = 143.43NM
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CHOOSE Ф:
FORCE APPLIED:50 NEWTONS
Distance:1.00 M
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90° 60° 45°
Г = (50 N)(1.00M)(sin??°)
Torque = ?? NM
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CHOOSE Ф:
FORCE APPLIED:50 NEWTONS
Distance:1.00 M
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90° 60° 45°
Г = (50 N)(1.00M)(sin45°)
Torque = 35.36 NM
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CHOOSE Ф:
FORCE APPLIED:50 NEWTONS
Distance:1.00 M
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90° 60° 45°
Г = (50 N)(1.00M)(sin60°)
Torque = 43.3 NM
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CHOOSE Ф:
FORCE APPLIED:50 NEWTONS
Distance:1.00 M
HOME
90° 60° 45°
Г = (50 N)(1.00M)(sin90°)
Torque = 50 NM
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CREDITS HOME
Mark Joseph Bantayan1900 Philippine Normal University, Taft Ave Manila
Images: http://www.google.com.ph
Quizzes: Sears, Zemansky, et. al., University Physics, 13th Ed. Halliday, Resnick. Fundamentals of Physics.
