PHY183-Lecture13

PHY183-Lecture13 - E x a m p le S t ill R in g s 1 Gymnast...

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Unformatted text preview: E x a m p le : S t ill R in g s ( 1 ) ! Gymnast of mass 55 kg hangs from still rings. ! Question 1: What is the tension in each rope? Physics for Scientists & Engineers 1 Fall Semester 2006 Lecture 13 Example September 14, 2006 Physics for Scientists&Engineers 1 1 September 14, 2006 Physics for Scientists&Engineers 1 2 E x a m p le : S t ill R in g s ( 2 ) ! Answer: no forces in x-direction; in y-direction: !F i y,i = T1 + T2 " mg = 0 ! Both ropes support gymnast equally ! Question 2: What changes if the rings do not hang straight down, but at an angle ! < 90° relative to the ceiling? E x a m p le : S t ill R in g s ( 3 ) ! ! y ! ! ! Equilibrium conditions: #F i x,i = T1 cos! " T2 cos! = 0 \$ T1 = T2 = T Example Example T1 = T2 ! T T + T ! mg = 0 " T = mg 1 2 "F i y,i = T1 sin ! + T2 sin ! # mg = 0 ! Combine both equations: 2T sin ! " mg = 0 # T = mg 2 sin ! x = 1 (55 kg) # (9.81 m/s 2 ) 2 = 270 N September 14, 2006 Physics for Scientists&Engineers 1 3 September 14, 2006 ! As angle gets smaller, T gets bigger! Physics for Scientists&Engineers 1 4 T w o P u lle y s ( 1 ) ! Mass m hung from a system of two pulleys T w o P u lle y s ( 2 ) ! First observation: tension in string is the same everywhere! ! Free-body diagram of block (in y direction): Fg ! T3 = 0 " T3 = Fg = mg ! Free-body diagram of pulley B above block (also y direction): 2T1 ! T3 = 0 " T1 = 1 T3 2 September 14, 2006 Physics for Scientists&Engineers 1 5 September 14, 2006 Physics for Scientists&Engineers 1 6 T w o P u lle y s ( 3 ) ! Combine these two equations: ! Final remark: T w o P u lle y s ( 4 ) • Combining all equations, we arrive at the conclusion: T1 = 1 T3 = 1 mg 2 2 ! => Important result: If you use this pulley system, you only need to apply a force equal to 1/2 of the weight of the object. ! The pulley A only serves to redirect the force T1 in this example: T2 = 2T1 = 2( 1 T3 ) = T3 = Fg 2 • This means that the ceiling mount of the upper pulley carries the entire weight of the block (compare green circles) • The entire force acting on the ceiling is exactly the same as the downward force on the rope, plus the weight of the block (see orange ovals) September 14, 2006 Physics for Scientists&Engineers 1 8 2T1 ! T2 = 0 " T2 = 2T1 September 14, 2006 Physics for Scientists&Engineers 1 7 F o r c e M u lt ip lie r ! We can also use the same rope in loops and guide it over the same pulley multiple times • In this example: 3 loops • We have six force vectors pointing up, all with string tension T • The object suspended from this pulley then can have a weight of 6T C e n t e r o f G r a v it y ! Where do we attach the gravity force vectors to the object in a free-body diagram? ! Center of Gravity • Exists for every object • Defined as the point where we can imagine the mass of an object to be concentrated • Objects with uniform mass density: center of gravity is at the geometrical center • More on center of gravity in chapter 8 • W e w ill le a r n h o w to c a lc u la te its lo c a tio n w ith th e a id o f s y m m e tr y c o n s id e r a tio n s a n d in te g r a l c a lc u lu s ! In general for n loops: T= September 14, 2006 1 mg 2n Physics for Scientists&Engineers 1 9 September 14, 2006 Physics for Scientists&Engineers 1 10 C e n t e r o f G r a v it y E x a m p le ! Again, holding the notebook computer Note: Marked the center of notebook via intersecting lines! Center of gravity C e n t e r o f G r a v it y ! General Principle: ! If an object (or group of objects) is not attached to other objects, it can only be stable, if its center of gravity is supported from directly below ! (Object can also be suspended from a string directly above the Center of Gravity) ! Fg ! Hand needs to support notebook below center of gravity - otherwise it will tip over! September 14, 2006 Physics for Scientists&Engineers 1 11 September 14, 2006 Physics for Scientists&Engineers 1 12 M o m e n t A rm ! If you want to tighten or loosen this bolt, which is the preferred way? T o rq u e ! Quantify this relationship: ! Definition of:Torque • Magnitude ! = rF sin " (a) (b) ! Pretty obvious: (c) is best ! What matters: (c) • Direction: clockwise or counter-clockwise • D is ta n c e to th e p iv o t p o in t • A n g le b e tw e e n fo r c e v e c to r a n d l in e c o n n e c tin g p iv o t p o in t a n d c o n ta c t p o in t o f fo r c e ( 9 0 ° o r 2 7 0 ° is b e s t; 0 ° a n d 1 8 0 ° is w o r s t) September 14, 2006 Physics for Scientists&Engineers 1 13 September 14, 2006 Physics for Scientists&Engineers 1 14 N e t T o rq u e ! Definition of net torque: sum of all clockwise torques minus sum of counterclockwise torques ! net = " ! clockwise, i # " ! counter-clockwise, j i j S t a t ic E q u ilib r iu m C o n d it io n 2 ! An object can only be in equilibrium, if the net torque acting on it is 0: ! net = 0 September 14, 2006 Physics for Scientists&Engineers 1 15 ...
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