1. AP Physics
Section 8-4
Torque

2. Torque
Torque is the rotational analog to linear force. A net torque produces an angular acceleration.
Torque is the tendency of a force to rotate an object around a axis, fulcrum, or pivot.
The symbol for torque is τ (Greek tau).
The units for torque are newton•meters (Nm).

3. Magnitude of torque is given by:
τ = r × F F
causes rotation
τ = r F F
τ = r F sinθ
doesn’t
cause
rotation
lever arm
torque up, out of paper
Direction of the torque vector ×
up, out of paper
down, into paper

4. The grip right hand rule
recall:
Clockwise displacement is considered negative.
Counterclockwise displacement is considered positive.
Net Torque
Στ = τcw + τccw
When there is no angular acceleration, Στ = 0.
This is called rotational equilibrium.

5. 1. Calculate the torques on the wrench. b.
0.20 m
131.4°
0.20 m
120 N
90° c.
120 N
τ = r F sinθ
41.4°
180°
0.15 m
τ = (0.20) (120N) sin90°
τ = (0.20) (120N) sin131.4°
τ = 24 Nm
τ = 18 Nm
τ = 0 Nm

7. 2. Calculate the net torque (magnitude and direction) on the beam due to the forces shown. The axis of rotation is perpendicular to the page.
F1 = 11 N, F2 = 23 N, and F3 = 15 N
If point 1 is pivot: 1 2 3
Torques about point 1
τ1 = r1 F1 sinθ
Since r1 = 0, τ1 = 0 Nm
τ2 = r2 F2 sinθ
τ2 = (3.0 m)(23 N)sin+90°
τ2 = +60 Nm (CCW)
τ3 = r3 F3 sinθ
τ3 = (4.0 m)(15 N)sin–45°
τ3 = –42 Nm (CW)
Στ = +18 Nm
Up out of page

8. If point 3 is pivot: Στ = –45 Nm Torques about point 3 τ1 = r1 F1 sinθ2 3
Torques about point 3
τ1 = r1 F1 sinθ
τ1 = (4.0 m)(11 N)sin–30°
τ1 = –22 Nm (CW)
τ2 = r2 F2 sinθ
τ2 = (1.0 m)(23 N)sin–90°
τ2 = –23 Nm (CW)
τ3 = r3 F3 sinθ
Since r3 = 0, τ3 = 0 Nm
Στ = –45 Nm
Down into page

9. τA + τB τcw = τcw = τccw = τA = rA FA sinθ τA = τB = rB FB sinθ τB =
24. Calculate the net torque about the axle of the wheel shown in the figure. A B C
Στ = τcw + τccw
Clockwise:
τA + τB
(0.12 m)(35 N) sin–90°
τA = rA FA sinθ =
τA =
–4.20 Nm
τB = rB FB sinθ
(0.24 m)(18 N) sin–90° =
τcw =
(–4.20)+(–4.32)
τB =
–4.32 Nm
τcw =
–8.52 Nm
Counterclockwise:
τC = rC FC sinθ
(0.24 m)(28 N) sin+90° =
τC =
τccw =
+6.72 Nm
+6.72 N
Στ = (–8.52 Nm) + (+6.72 Nm) =
–1.8 Nm