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Unformatted text preview: September 25, 2003 Physics 16 Assignment #1 11. l 1 l 1 l 2 We can solve this problem by focusing on the l 1 segment on the left of the diagram, which is redrawn here as a part of a triangle the other two segments of which come from the red segments in the above drawing. The force vector F t here has opposite direction to its equivalent in the problem, but the same magnitude. F s is also shown in the opposite direction, and broken down into its horizontal and vertical components, F h and F v . F t m 1 ge l 1 F h e F ve 1 e Note that F t must be entirely horizontal. To see this, consider that the vertical component of the force that the first l 1 rod exerts on the second l 1 rod must, by Newton's third law, be equal in magnitude but opposite in direction to that the second rod exerts on the first, i.e., F l1>l2 = F l2>l2 . But, by symmetry, both the direction and the magnitude of those forces must be equal, i.e., F l1>l2 = F l2>l2 . This can be true only if F l1>l2 =F l2>l2 = 0. Since the tetrahedron is static, the horizontal and vertical forces must each sum to zero: F h = F t m 1 g = F v Likewise, the torques must all sum to zero. Taking torques around the bottom left pivot (clockwise positive) gives: T = m 1 g*l 1 /2*cos(e)  F t l 1 sin(e) = 0 To find e, we first find the length of the base of the above triangle. We can do this by looking at a top view of the base of the tetrahedron with its medians drawn in: l 2 l 2 l 2 The dashed lines each have length (l 2 e3)/2, and the base of the triangle in the freebody diagram has length 2/3 of the dashed line, or (l 2 e3)/3. Going back to that triangle, we have: l 1 2 3 1 1 2 2 l l (l 2 e3)/3 This gives: 2 e 1 3 2 1 1 ] sin[ 2 2 l l = θ 1 2 3 3 ] cos[ l l = θ Now we can simplify our original torque equation by plugging in the values for sin(e) and cos(e): m 1 gcos(e)l 1 /2 = F t l 1 sin(e) 1 3 2 1 1 6 3 2 2 1 2 1 l l t F l l g m = 2 2 2 1 2 1 2 2 1 2 1 3 2 1 3 2 1 1 6 3 l l gl m l gl m F l l t = = 2 4 1 12 2 3 1 12 ) ( 2 2 2 2 1 2 2 1 l l l l F gm t g m F s = + = Dimensional analysis shows that both answers have the correct units: F t = (kg(m/s 2 )m)/m = N F s = (m/s 2 )kg = N Taking limiting values as l 1 or l 2 goes to a minimum value also gives expected results. As l 2 goes to zero (l 2 << l 1 ), the tetrahedron degenerates into three vertical rods, so F t goes to zero and F s goes to mg. As l 1 goes to its minimum possible value (l 1> (l 2 e3)/3), it becomes increasingly difficult to supply the necessary torque to keep the rods from falling, and both F t and F s approach infinity. 3 12. First, we solve the given differential equation to find V[t]. β α ) ( ) ( ' t V t m V = d m t V d V B α = ) ( ' ) ( ) ( ∫ ∫ = t t t V t V dt m V dV α β [ ] ) ( ) ( ) ( 1 1 1 1 t t m t V t V = α β β β β β β α + = 1 1 1 ) ( ) 1 )( ( ) ( t V t t m t V Then, we can integrate V(t) to find X(t): ∫ ∫ + = t t t...
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This note was uploaded on 09/08/2008 for the course PHYS 16 taught by Professor Staff during the Spring '08 term at Harvard.
 Spring '08
 Staff
 Physics

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