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### PHY183-Lecture30pre

Course: PHY 183, Spring 2012
School: Michigan State University
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Word Count: 1574

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Conditions Equilibrium In chapter 4 we explored the conditions for mechanical equilibrium We found that the necessary condition for mechanical equilibrium is the absence of a net external force In that case Newtons First Law stipulates that an object in mechanical equilibrium stays at rest or moves with constant velocity But most often we want to explore the conditions that are needed for an object to stay at...

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Conditions Equilibrium In chapter 4 we explored the conditions for mechanical equilibrium We found that the necessary condition for mechanical equilibrium is the absence of a net external force In that case Newtons First Law stipulates that an object in mechanical equilibrium stays at rest or moves with constant velocity But most often we want to explore the conditions that are needed for an object to stay at rest, in static equilibrium Physics for Scientists & Engineers 1 Spring Semester 2012 Lecture 30 Static Equilibrium March 20, 2012 Physics for Scientists&Engineers 1 1 March 20, 2012 The requirement of no translational or rotational motion means that the linear and angular velocities of this object are zero at all times Since the values of the linear and angular velocities do not change in time A famous example for such a collection of objects in static equilibrium is shown below Part of what makes this sculpture by Alexander Calder so amazing is that the eye does not want to accept that the configuration it is in is a stable static equilibrium Physics for Scientists&Engineers 1 The linear and angular accelerations are zero at all times Fnet = ma = 0 and net = I = 0 Static Equilibrium Condition 1: An object can stay in static equilibrium only if the net force acting on it is zero. Static Equilibrium Condition 2: An object can stay in static equilibrium only if the net torque acting on it is zero. Even if Newtons First Law is satisfied and an object has no translational motion, it will still rotate if it experiences a net torque. 3 March 20, 2012 Equilibrium Conditions 4 If an object is supported from a pin located directly above its center of mass, then the object stays balanced It does not move or rotate For static equilibrium with zero net torque, the net torque has to be zero for any pivot point chosen A clever choice of pivot point can be the key to a quick solution For example, we can select the point where an unknown force acts as the pivot point Then the unknown force would not contribute to the net torque because it would have a zero moment arm Physics for Scientists&Engineers 1 Physics for Scientists&Engineers 1 Equilibrium Conditions Torque is always defined with respect to a pivot point The pivot point is the intersection of the plane determined by the force and moment arm vectors and the axis of rotation When we compute the net torque, the pivot point must be the same for all forces March 20, 2012 2 Equilibrium Conditions Equilibrium Conditions March 20, 2012 Physics for Scientists&Engineers 1 5 Why? The only two forces acting on the object are the force of gravity and the normal force The two forces cancel each other and produce no torque If the object is supported in the same way from a pin but its center of mass is not below the support point, a net torque acts Object is not in static equilibrium March 20, 2012 Physics for Scientists&Engineers 1 6 1 Experimentally Experimentally Determining the Center of Mass Experimentally Determining the Center of Mass We can locate the center mass for many objects supporting the object on two fingers such that the center of mass is located somewhere between them Slowly slide the fingers closer to each other The finger closer to the center of mass exerts a higher normal force and thus a high friction force When the fingers meet, you have determined the center of mass We can locate the center of mass of an object experimentally by suspending the object from different points First, support the object in such a way that it can rotate freely Use a plumb bob to mark the vertical line Support the object from a second point such that it can rotate freely and mark the vertical line The center of mass will lie at the intersection of the lines March 20, 2012 Physics for Scientists&Engineers 1 7 March 20, 2012 Equilibrium Equations Physics for Scientists&Engineers 1 8 Seesaw The equations for equilibrium include zero net force The seesaw is a playground toy, which consists of a pivot and a bar, of mass M, that is placed over it so that its ends can move up or down freely as shown below n Fnet, x = Fi , x = F1, x + F2, x + ... + Fn, x = 0 i =1 n Fnet, y = Fi , y = F1, y + F2, y + ... + Fn, y = 0 i =1 And zero net torque net = counter-clockwise,i clockwise, j = 0 i March 20, 2012 j Physics for Scientists&Engineers 1 9 March 20, 2012 Seesaw Physics for Scientists&Engineers 1 10 Seesaw If you place a an object of mass m1 on one end at a distance r1 from the center axle, then obviously that end goes down, simply because of the force and torque that the object exerts on it Question 1: Where do you have to place an object of mass m2 to get the see saw to be balanced, so that the bar is horizontal and neither touches the end ground? March 20, 2012 Physics for Scientists&Engineers 1 11 The force that m1 exerts on the bar is simply m1g, acting downward The same is true for force that m2 exerts on the bar The weight of the bar is Mg acting at the center of mass of the bar The final force is the normal force N exerted by the support of the seesaw acting on the axle of the seesaw March 20, 2012 Physics for Scientists&Engineers 1 12 2 Seesaw Standing on a board Our equilibrium force equation is A construction supervisor of mass M = 90.6 kg is standing on a board of mass m = 27.0 kg. Two sawhorses at a distance l = 4.40 m apart support the board. If the man stands a distance x1 = 1.188 m away from the left-hand sawhorse as shown, what is the magnitude of the force that the board exerts on that sawhorse? Fnet, y = Fi , y = m1g m2 g Mg + N = 0 i N = g( m1 + m2 + M ) Our equilibrium torque equation is net = clockwise,i counter-clockwise, j i j = m2 gr2 sin 90 m1gr1 sin 90 = 0 m2 r2 = m1r1 m r2 = r1 1 m2 March 20, 2012 Physics for Scientists&Engineers 1 13 Standing on a board FR pivot mg clockwise, j = FL 15 counter-clockwise,i i net = counter-clockwise,i clockwise, j = 0 i March 20, 2012 Stacking Blocks = mg + Mg x2 2 j mg + Mg x2 FL = 0 2 mg + Mg x2 FL = 0 2 Mg x2 + mg 2 FL = 90.6kg 9.81 Physics for Scientists&Engineers 1 j FL = Mg March 20, 2012 14 Standing on a board Find the force on the board exerted by the left sawhorse Draw a free-body diagram The net torque is zero net = counter-clockwise,i clockwise, j = 0 i j Choose a pivot point FL Physics for Scientists&Engineers 1 Calculate the torques M = 90.6 kg, m = 27.0 kg. l = x1 + x2 = 4.40 m and x1 = 1.188 March 20, 2012 FR FL mg pivot Mg m m 4.4m [ 4.4 1.188] m + 27.0kg 9.81 2 s2 s2 = 781N 4.4m Physics for Scientists&Engineers 1 16 Stacking Blocks Start with one block If the block has length and uniform mass density, then it center of mass is located at (1/2) We can use the definition of the center of gravity to produce a surprising result Lets start with a stack of seven identical blocks Surprising result: Last block is completely over the edge of the table! How far can we push out the leading edge of the block without the pile of blocks falling over? March 20, 2012 Physics for Scientists&Engineers 1 We call the x-coordinate of the center of mass x1 17 March 20, 2012 Physics for Scientists&Engineers 1 18 3 Stacking Blocks Stacking Blocks We call the x-coordinate of the center of mass of the second block x2 The center of mass of the combined system of two blocks is xm +x m x12 = 1 1 2 2 m1 = m2 x12 = 1 ( x1 + x2 ) 2 m1 + m2 Block 1 can stay at rest as long as at least half of it is supported from below and can stick out an infinitesimal amount less than (1/2) and remain at rest Now add a second identical block In the limiting case that the center of mass of the first block is still supported from below by the second block x1 = x2 + 1 2 Combining these two equations gives us x12 = 1 ( x1 + x2 ) = 2 x12 = 1 (( 1 + x2 ) + x2 ) 2 2 x12 = x2 + 1 4 March 20, 2012 Physics for Scientists&Engineers 1 19 March 20, 2012 Physics for Scientists&Engineers 1 Stacking Blocks 20 Stacking Blocks Shifting x12 to the very edge of the third block gives us Now add a third block x12 = x3 + 1 2 Combining this with our result for two blocks, we get x12 = x3 + 1 = x3 + 1 x2 = x3 + 1 4 4 2 The center of mass of the three blocks is x12 ( 2 m ) + x3 m 2m + m x123 = 2 x12 + 1 x3 3 3 x123 = The top two blocks will not topple if the combined center of mass of x12 is supported from below March 20, 2012 Physics for Scientists&Engineers 1 21 x123 = 2 3 ( x3 + 1 ) + 13 x3 2 x123 = x3 + 1 3 March 20, 2012 Stacking Blocks Physics for Scientists&Engineers 1 22 Stacking Blocks Now add a fourth block and require that the top three blocks are supported from below Combining our equations so far we get x123 = x3 + 1 = x4 + 1 x3 = x4 + 1 3 2 6 x123 = x4 + 1 2 We could keep going on to the fifth block, but lets think If we have (n 1) blocks supported by the nth block, then the coordinates of the (n 1)st and nth block are related as xn 1 = xn + 2n 2 We can add up all the terms and find out how far away x1 can be from the edge of the table n x1 = xn +1 + ( 1 ) 2 i =1 March 20, 2012 Physics for Scientists&Engineers 1 23 March 20, 2012 1 i Physics for Scientists&Engineers 1 24 4 Stacking Blocks We get the amazing result that x1 can move infinitely far from the edge of the table Here are our seven blocks The last block is completely over the edge March 20, 2012 Physics for Scientists&Engineers 1 25 5
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