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Torque

Kinetic art by Anthony Howe, 2010

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Torque, τ

  • Force is the action that creates �changes in linear motion.
  • For rotational motion, the same force can cause very different results.
  • A torque is an action that causes objects to rotate.
  • A torque is required to rotate an object, just as a force is required to move an object in a line.

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Location, location, location!

  • Torque is created by force, but it also depends on where the force is applied and the point about which the object rotates.
  • For example, a door pushed at its handle will easily turn and open, but a door pushed near its hinges will not move as easily. The force may be the same but the torque is quite different.

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Center of rotation

  • Center of rotation: the point or line about which an object turns.
    • For example, a door’s center of rotation is at its hinges.
  • A force applied far from the center of rotation produces a greater torque than a force applied close to the center of rotation.
  • What happens when a force is applied at the hinges?

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Line of action

  • Line of action: the location and direction of the force causing the torque.

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Force applied must be perpendicular

  • The lever arm, r (also called a moment arm) is the perpendicular distance between the line of action of the force and the center of rotation

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Calculating torque�

The torque (τ) created by a force is equal to the lever arm length (r) times the magnitude of the force (F).

  • The units of torque are force times distance, or newton·meters (N·m)
  • Forces applied at the center of rotation create zero torque.

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Torques can be combined

  •  

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Net torque practice

τnet = τA + τB

= FArA + FBrB

= +(10N)(.2m) + -(10N)(.1m)

= 1 N·m

What is the net torque acting on the freely rotating object shown below.

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You try…

  • Note: When considering the torque exerted by the mass of the lever itself, treat the mass of the lever as if it is concentrated at its center �of mass.

A force of 50 newtons is applied to a 0.50 kg wrench that is 30 centimeters long. Calculate the net torque acting on the wrench if the applied force is perpendicular to the wrench at the end of the wrench.

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You try…

A force of 50 newtons is applied to a 0.50 kg wrench that is 30 centimeters long. Calculate the net torque acting on the wrench if the applied force is perpendicular to the wrench at the end of the wrench.

τnet = τapplied + τwrench mass

= Fappliedrapplied + FwrenchrCMwrench

= -(50 N)(0.3m) + -(.5kg)(9.8m/s2)(.15 m)

= -16 Nm

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Statics

Any object that is not moving is in

    • rotational equilibrium �(τnet = 0 Nm)

and

    • translational equilibrium �(Fnet = 0 N)

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Example

  • Consider a 10-meter bridge that weighs 500 newtons supported at both ends. A person who weighs 750 newtons is standing 2 meters from one end of the bridge.
  • What are the forces (FA, FB) holding the bridge up at either end?

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Example

Translational equilibrium

Rotational equilibrium

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Example

Translational equilibrium

Rotational equilibrium

The bridge is not moving up or down. Fnet = 0 N

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Example

Translational equilibrium

Rotational equilibrium

The bridge is not moving up or down. Fnet = 0 N

The bridge is not rotating.

τnet = 0 Nm

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Example

Translational equilibrium

Rotational equilibrium

The bridge is not moving up or down. Fnet = 0 N

The bridge is not rotating.

τnet = 0 Nm

Fnet = FA + FB + Fperson + Fbridge

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Example

Translational equilibrium

Rotational equilibrium

The bridge is not moving up or down. Fnet = 0 N

The bridge is not rotating.

τnet = 0 Nm

Fnet = FA + FB + Fperson + Fbridge

τnet = τA + τB + τbridge + τperson

Any point can be considered the center of rotation since the bridge is static, so pick a point that eliminates one of the torques, such as point A. So now, τA = 0 Nm.

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Example

Translational equilibrium

Rotational equilibrium

The bridge is not moving up or down. Fnet = 0 N

The bridge is not rotating.

τnet = 0 Nm

Fnet = FA + FB + Fperson + Fbridge

τnet = τA + τperson + τbridgeB

0 Nm = 0 Nm + -(750Nx2m) + � -(500Nx5m) + (FBx10m)

Any point can be considered the center of rotation since the bridge is static, so pick a point that eliminates one of the torques, such as point A. So now, τA = 0 Nm.

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Example

Translational equilibrium

Rotational equilibrium

The bridge is not moving up or down. Fnet = 0 N

The bridge is not rotating.

τnet = 0 Nm

Fnet = FA + FB + Fperson + Fbridge

τnet = τA + τperson + τbridgeB

0 Nm = 0 Nm + -(750Nx2m) + � -(500Nx5m) + (FBx10m)

FB = 400 N

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Example

Translational equilibrium

Rotational equilibrium

The bridge is not moving up or down. Fnet = 0 N

The bridge is not rotating.

τnet = 0 Nm

Fnet = FA + FB + Fperson + Fbridge

τnet = τA + τperson + τbridgeB

0 N = FA + 400N + -750 N + -500 N

0 Nm = 0 Nm + -(750Nx2m) + � -(500Nx5m) + (FBx10m)

FB = 400 N

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Example

Translational equilibrium

Rotational equilibrium

The bridge is not moving up or down. Fnet = 0 N

The bridge is not rotating.

τnet = 0 Nm

Fnet = FA + FB + Fperson + Fbridge

τnet = τA + τperson + τbridgeB

0 N = FA + 400N + -750 N + -500 N

0 Nm = 0 Nm + -(750Nx2m) + � -(500Nx5m) + (FBx10m)

FA = 850 N

FB = 400 N

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Force and lever arm are not always�perpendicular�

When the force and lever arm are not perpendicular, an extra step is required to calculate torque.

Original equation: τ = rF

New equation: τ = rF or

τ = rFsinθ

θ

θ = the angle from the lever arm to the line of action

r

F

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You try…

  • A 20-cm wrench with a mass of 0.50 kg is used to loosen a bolt. A force is applied 0.20 m from the bolt at an angle of 30o from the perpendicular. If it takes 50 N to loosen the bolt when the force is applied perpendicularly to the wrench, how much force must be applied at the angle to loosen the bolt?

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Three forces labeled A, B, C are applied to a rod which pivots on an axis through its center.

Which force causes the largest magnitude torque?

  1. A C) C
  2. B D) two or more forces tie for largest size torque.

AP Multiple Choice Practice

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A door is pushed on by two forces, a smaller force at the door knob and a larger force nearer the hinge as shown. The door does not move. The force exerted on the door by the hinge ____.

A) is zero C) points ↓ (along -y)

B) points (along +y) D) points (lower right, in diagram)

AP Multiple Choice Practice

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What is the ratio ?

(Hint: consider the torque about the mass M).

A) 2/3 B) 1/3 C) 1/2 D) 2

A mass M is placed on a very light board supported at the ends, as shown. The free-body diagram shows directions of the forces, but not their correct relative sizes.

AP Multiple Choice Practice