Torque
Kinetic art by Anthony Howe, 2010
Torque, τ
Location, location, location!
Center of rotation
Line of action
Force applied must be perpendicular
Calculating torque�
The torque (τ) created by a force is equal to the lever arm length (r) times the magnitude of the force (F).
Torques can be combined
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.
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.
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
Statics
Any object that is not moving is in
and
Example
Example
Translational equilibrium | Rotational equilibrium |
| |
Example
Translational equilibrium | Rotational equilibrium |
The bridge is not moving up or down. Fnet = 0 N | |
Example
Translational equilibrium | Rotational equilibrium |
The bridge is not moving up or down. Fnet = 0 N | The bridge is not rotating. τnet = 0 Nm |
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 | |
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.
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 + τbridge +τB |
| 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.
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 + τbridge +τB |
| 0 Nm = 0 Nm + -(750Nx2m) + � -(500Nx5m) + (FBx10m) |
| FB = 400 N |
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 + τbridge +τB |
0 N = FA + 400N + -750 N + -500 N | 0 Nm = 0 Nm + -(750Nx2m) + � -(500Nx5m) + (FBx10m) |
| FB = 400 N |
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 + τbridge +τB |
0 N = FA + 400N + -750 N + -500 N | 0 Nm = 0 Nm + -(750Nx2m) + � -(500Nx5m) + (FBx10m) |
FA = 850 N | FB = 400 N |
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 = r┴F or
τ = rFsinθ
θ
θ = the angle from the lever arm to the line of action
r
F
You try…
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?
AP Multiple Choice Practice
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
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