_lec11_ppt_ - Lecture 11 P112 Announcements Prelim 1 on Mar 1 HW 4 due on Friday Lab 3 this week Lab 4 next week Want to choose the pre-lecture

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Unformatted text preview: Lecture 11 P112 Feb 14, 2007 Announcements Prelim 1 on Mar 1 HW 4 due on Friday Lab 3 this week! Lab 4 next week Want to choose the pre-lecture music? Bring up your music player before class. Agenda for today Tools: Springs, Ropes, Pulleys Statics Tools: Pulleys Used to change the direction of forces An ideal massless pulley will change the direction of an applied force without altering the magnitude: F1 ideal peg or pulley F2 | F1 | = | F2 | Tools: Ropes & Strings... Consider a horizontal segment of rope with mass m: m T1 a T2 x FBD: T1 T2 WEonR Tools: Ropes & Strings... Using Newton's 2nd law (in x direction): Fx = T2 - T1 = ma So if m = 0 or/and a = 0 then T1 = T2 An ideal (massless) rope has constant tension along the rope Will mostly deal with massless ropes in this course! Tools: Ropes & Strings... If a rope has mass, the tension can vary along the rope For example, a heavy rope hanging from the ceiling... T = mg T=0 Quick problem: Ideal pulley and massless rope: in which situation is the rope more likely to break? 1) A 2) B 3) both the same A B Springs Hooke s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. FX = -k x x: displacement from relaxed position k: spring constant Springs relaxed position FX = 0 x x=0 FX = -kx > 0 x<0 FX = - kx < 0 x>0 Statics Statics Conditions for particles in equilibrium Either at rest or moving at constant velocity r F =0 Condition (N1): In component form: Fx = 0 Fy = 0 Problem 1: Tension in Ropes A box is suspended from the ceiling by two ropes making an angle with the horizontal. What is the tension in each rope? m Solution Draw a FBD: T1sin T1cos T2cos W Tr1onB Tr2onB T2sin Since the box is in equilibrium, Fx = 0 and Fy = 0 Fx,NET = T1cos Fy,NET = T1sin - T2cos + T2sin =0 - mg = 0 T1 = T 2 mg T1 = T2 = 2 sin Problem 2:Block and Spring A block of mass M = 5.1 kg is supported on a frictionless ramp by a spring (k = 125 N/m). When the ramp is horizontal the equilibrium position of the mass is at x = 0. When the angle of the ramp is changed to 30o what is the new equilibrium position of the block x1? =? x1 k x=0 M k M = 30o Solution Choose the x-axis to be along direction of ramp. NRonB FSonB WEonB In equilibrium: Fx=0 forces in x direction: FSonB = kx W x = sin W EonB y x 5.1kg 9.81m s2 0.5 x1 = = 0.2m 125 N m Recap Tools: Ideal pulleys redirect forces Ideal massless ropes have constant tension along the rope Ideal Springs: Hooke s law Statics: use the fact that the x and y components of the net force is zero in equilibrium as a constraint to calculate tensions, positions,... Choose Axes wisely Next time: dynamics! ...
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell University (Engineering School).

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