Lect17

# Ladders sign posts balanced beams buildings bridges r

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Unformatted text preview: systems that don’t move. Ladders, sign-posts, balanced beams, buildings, bridges... r r The key equations are familiar to us: ∑ F = ma r r ∑τ = Iα r r If an object doesn’t move: a = 0 τ = 0 If r ∑ F = 0 The net force on the object is zero r The ∑ τ = 0 The net torque on the object is zero (for any axis) Physics 211 Lecture 17, Slide 4 Statics: Statics: Example: What are all of the forces acting on a car parked on a hill? What y x N f θ mg Physics 211 Lecture 17, Slide 5 Physics 211 Lecture 17, Slide Car on Hill: Car on Hill: • Use Newton’s 2nd Law: FNET = MACM = 0 Use NET CM • Resolve this into x and y components: Resolve x: f - mg sin θ = 0 f = mg sin θ y: N - mg cos θ = 0 y x N f N = mg cos θ θ mg Physics 211 Lecture 17, Slide 6 Physics 211 Lecture 17, Slide Torque due to Gravity Torque due to Gravity Magnitude: R⊥ r RCM τ = Rcm Mg sin(θ ) = MgR⊥ Lever arm Physics 211 Lecture 17, Slide 7 Example: Example: Now consider a plank of mass M suspended by two strings as Now shown. We want to find the tension in each string: shown. T1 T2 M x cm L/4 L/2 Mg y x Physics 211 Lecture 17, Slide 8 Balance forces: Balance forces: r ∑F = 0 T1 + T2 = Mg T1 T2 M x cm L/4 L/2 Mg y x Physics 211 Lecture 17, Slide 9 Balance torques r ∑τ = 0 Choose the rotation axis to be out of the page through the CM: page The torque due to the string on the right about this axis is: L τ 2 = T2 4 The torque due to the string on the left about this axis is: L τ1 = −T1 2 T1 T2 M x cm L/4 L/2 Mg y Gravity exerts no torque about the CM x Physics 211 Lecture 17, Slide 10 Finish the problem The sum of all torques must be 0: L L −T1 + T2 = 0 2 4 τ1 + τ 2 = 0 T 2 = 2T1 T1 M x cm We already found that T1 + T2 = Mg T1 = T2...
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