This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 = 1 2 3 3 ] cos[ l l = 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]. ) ( ) ( ' t V t m V = d m t V d V B  = ) ( ' ) ( ) (  = t t t V t V dt m V dV [ ] ) ( ) ( ) ( 1 1 1 1 t t m t V t V =  + = 1 1 1 ) ( ) 1 )( ( ) ( t V t t m t V Then, we can integrate V(t) to find X(t):  + = t t t...
View Full
Document
 Spring '08
 Staff
 Physics

Click to edit the document details