This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: we have: F N – mg + dm/dt v rel y = 0 In order to find F N we need to have an expression for dm/dt. Let’s first find the relation between dm and dl , where dm is the mass of a short rope segment of length dl that falls on the scale during time dt . Since the rope is uniform, the relation between dm and dl is dm/dl = M / L and multiplying both side for dt dm/dt = M/ L dl/dt = M/L v rel y Note that d/dt is the impact speed of the segment, so v rel y =dl/dt (v rel y is negative because up is the positive y direction and the rope is falling). Substituting in the equation for the force we have: F N = mg + M/L v 2 rel y To find v 2 rel y you need to note that, until it touches the scale, each point along the rope falls with constant acceleration g, therefore using the usual equation for a motion with constant acceleration we have: v 2 rel y = v 2 rel 0 +2a y Δ y = 0 + 2(g) (L/2) = gL and substituting in F N = Mg/2 + MgL/L = 3/2 Mg where m=M/2...
View
Full Document
 Spring '08
 Lanzara
 Physics, Force, Mass, General Relativity, Fundamental physics concepts, 2 mg

Click to edit the document details