Unit 5, Lesson 5 Torque

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...

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 “–”

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° = -37.5 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. Find the net torque about a) point A b) point B TA = +(200)(2)sin 60 – (150)(4) sin 60 = - 173 Nm TB = +(100)(2) – (150)(2)sin 60 = - 59.8 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

Rotational Equilibrium Unit 5, Lesson 6 Rotational Equilibrium

Conditions for Equilibrium, revisited ∑ F = 0 (Translational) ∑ T = 0 (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.

Step 2 Choose a pivot point, use F1 Step 1 Draw a FBD F2 1.5m F1 4.0m 800 N Step 2 Choose a pivot point, use F1 Step 3 Clockwise torques = counter clockwise torques (note: no torques at F1 at this point) 800 x 4 = F2 x 1.5 F2 = = 2130 N Step 4 Use net force = 0 to find the other force F2 = F1 + 800 F1 = F2 – 800 = 2130 – 800 = 1330 N

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 1.5 + 100 x 2.5 = F2 x 3 550 = F2 x 3 F2 = 183 N F1 + 183.3 = 200 + 100 F1 = 117 N Give FBD to start them off – 15 min

Brain Break!

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

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

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

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

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 3 + 600sin30 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 = 300+600 Hy = 796 N

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

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

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