You are lowering two boxes, masses m, and m,, one on the top of the other, down the ramp by pulling ona rope parallel to the surface of the ramp. The coefficient of friction between the blocks is u̟, and thecoefficient of friction between the lower block and the ramp is uz. The pulling force of magnitude P issuch that they are moving with constant velocity. The angle 0 is known.a) Draw the free body diagram for both blocks.b) In the box below write the system of equations that could be soled to find force P and the force offriction between the two blocks.c) Find the magnitude of the force P and magnitude of the force of friction between the blocks.m,

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Asked Dec 9, 2019
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You are lowering two boxes, masses m, and m,, one on the top of the other, down the ramp by pulling on
a rope parallel to the surface of the ramp. The coefficient of friction between the blocks is u̟, and the
coefficient of friction between the lower block and the ramp is uz. The pulling force of magnitude P is
such that they are moving with constant velocity. The angle 0 is known.
a) Draw the free body diagram for both blocks.
b) In the box below write the system of equations that could be soled to find force P and the force of
friction between the two blocks.
c) Find the magnitude of the force P and magnitude of the force of friction between the blocks.
m,
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You are lowering two boxes, masses m, and m,, one on the top of the other, down the ramp by pulling on a rope parallel to the surface of the ramp. The coefficient of friction between the blocks is u̟, and the coefficient of friction between the lower block and the ramp is uz. The pulling force of magnitude P is such that they are moving with constant velocity. The angle 0 is known. a) Draw the free body diagram for both blocks. b) In the box below write the system of equations that could be soled to find force P and the force of friction between the two blocks. c) Find the magnitude of the force P and magnitude of the force of friction between the blocks. m,

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Expert Answer

Step 1
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Part (a) mg mi mgroso mig Vmg Free Bodly Diagrom for free body diagram for lowen upper block.. bloek FIGORE-d FIGURE -2

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Step 2
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Îo N=(mit Me)gcoso (mitmz)g 18 Free body diagram for the two blocks FIQURE (3)

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Step 3
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Parts (Bland (c) From figure (3), E f, =0 sinre the blocks are mouing with constant velocity. + (m,tm)gsino = P+f P= (mitmz) gsino - { P- (mit my) grino - MzN P= (mitM2) gsino - Mz (Mitma)gcoso p= (mitm2]g [sino - Hz cOso)

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