1. Unit 5, Lesson 5
Torque

2. Conditions for Equilibrium:
The object doesn’t move.
vector
∑ F = 0
Sum of
10N
eg)
10N
∑ F = 0
but the object moves! (rotates)
This object is in translational but not rotational equilibrium.
To understand rotational equilibrium, we must first learn about torque...

3. Torque (twisting force)
Simple case d
hinge
Torque = Force x distance (Newton-meters)
T = F ∙ d ( note F d ) T T F
General case F T d
) θ
θ = angle between beam & force
hinge
T = F ∙ d = F(sinθ) ∙ d T
Sign of Torque: counter clockwise “+”
clockwise “–”

4. Calculate the Torque 1. θ = 30° T = -F ∙ d sinθ = (-30)(2.5)sin30°
2.5m
60°
θ = 30°
T = -F ∙ d sinθ
= (-30)(2.5)sin30°
= Nm
30° (
3.0m
7.0N
) 50° 2.
θ = 50°
T = F ∙ d sinθ
= (7.0)(3.0)sin50°
= 16.1Nm
) 30°
20N
1.5m 3.
θ = 60°
T = -F ∙ d sinθ
= -(20)(1.5)sin60°
= -26Nm
3.0m
5.0m
20N
Demo #1-2, then have students attempt #3-4 – 30 min
) θ 4.
θ =
= 31° d d=
= 5.8 m
T = -F ∙ d sinθ
= (-5.8)(20)sin31°
= -60Nm
) θ

5. 5. Find the net torque about a) point A b) point B
TA = +(200)(2)sin 60 – (150)(4) sin 60
= Nm
TB = +(100)(2) – (150)(2)sin 60
= Nm 2m
100 N
200 N
150 N
60° ( A B
Tune in next class to learn about rotational equilibrium...
Students attempt #5 – 20 min

6. Rotational Equilibrium
Unit 5, Lesson 6
Rotational Equilibrium

7. Conditions for Equilibrium, revisited
∑ F = (Translational)
∑ T = (Rotational)
That is, clockwise torques must balance counter clockwise torques.
Examples
800 N
Diver 1.
2.5m
5 min
massless board
1.5m
Find the force exerted by each support.

8. Step 2 Choose a pivot point, use F1
Step 1 Draw a FBD F2
1.5m F1
4.0m
800 N
Step Choose a pivot point, use F1
Step Clockwise torques = counter clockwise torques
(note: no torques at F1 at this point)
800 x 4 = F2 x 1.5
F2 =
= 2130 N
Step Use net force = 0 to find the other force
F2 = F
F1 = F2 – 800
= 2130 – 800
= 1330 N

9. Find the force exerted by each support on the uniform beam.2.
200 N
Find the force exerted by each support on the uniform beam.
2.5 m
0.5 m F1 F2
FBD
2.5 m
1.5 m
200 N
100 N
Pivot point at
cw = ccw F1
200 x x 2.5 = F2 x 3
550 = F2 x 3
F2 = 183 N
F =
F1 = 117 N
Give FBD to start them off – 15 min

10. Brain Break!

11. Torque with Angles! 1.
Find the tension in the wire, the beam is uniform.
500 N
35° (
3.0 m
wire
500 N
35° ( T
1.5m
FBD
cw = ccw
(500)(1.5) = T (sin35)(3)
T = 436 N
∑ T = 0
(T sinθ)(3) – (500)(1.5) = 0 or
wire
40° (
4.0 m 2.
A uniform beam of weight 20 N and 4.0 m long is supported by a wire.
Find tension in the wire.
Find the force exerted by the hinge on the beam.
I do #1 (5 min), you do #2a (15 min) T
40° ( a)
cw = ccw
T (sin 40)(4)= (20)(sin 50)(2)
T =11.9N
FBD H
mg=20N

12. T 20N b) FBD X: Hx = T = 11.9N Y: Hy = mg = 20N H = =23.2Nθ Hx Hy b)
FBD X:
Hx = T = 11.9N Y:
Hy = mg = 20N
H = =23.2N
w/ horizontal
I do – 5 min

13. More Rotational Equilibrium!
Unit 5, Lesson 7
More Rotational Equilibrium!

14. The force exerted by the wall on the beam rope
3. The 200 N beam is 5.0 m long. Find
The tension in the rope
The force exerted by the wall on the beam
30° (
rope
100N 2m
200 N
30° ( T
100 N 5m a)
T(sin30)(5)=(200)(2.5)+(100)(2)
T = 280N
We do – 20 min b)
30° ( T W Wx Wy X:
Wx = T cos30 = 242.5N Y:
280 sin30 + Wy = 300 N
Wy = 160 N
W = 290 N

15. A uniform beam of weight 300 N supports a 600 N weight as shown. Find
wire
4.0 m
600N
70° ( 2m
) 60° 4.
A uniform beam of weight 300 N supports a 600 N weight as shown. Find
The tension in the wire.
The components of the force exerted by the hinge. T
300N H
) 60°
10° ( a)
600N a)
300sin30 x sin30 x 6 = Tsin70 x 4
T = 599 N
FBD T
900N θ Hx Hy
10° ( H b) X:
Hx = 599cos10 = 590 N
FBD
We do – 20 min Y:
Hy + 599sin10 =
Hy = 796 N

16. 5. A uniform 25 kg ladder 3.0 m long leans against a frictionless wall at an angle of 20O. Find the minimum coefficient of friction between ladder and ground.
ΣFx = 0
W = f
ΣFy = 0
N = mg = 25x9.8 = 245 N
Σ τ = 0 (bottom as pivot)
W(3.0)sin70 = (245)(1.5)sin20
W = 245sin20/(2sin70) = 44.6 N
µ = f/N = 44.6/245 = .18
We do - 20 min

17. Lesson 8: Practice with Torque
Unit 5: Equilibrium
Lesson 8: Practice with Torque

18. Practice Time!
You have today’s class to work on the Torque Problem Set.
60 min