Question

Two masses m1 and m2 are connected by a light string passing over a smooth pulley. When set free m1 moves downward by 2m in 2sec. Find ratio m1/m2 (the answer is 11\9) but help me in solving it

Answer 1

Two masses, M1 and M2 are attached to a massless string that is hung over a massless and frictionless pulley. When released from rest, the two masses move with acceleration a = 0.9 g in a clockwise direction. What is the ratio M1/M2 equal to?

Answer

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3 ANSWERS

Don Rolph, lives in East Walpole, MA

Answered Nov 26, 2015

Ok assumptions:

1) string massless

2) no friction in pulley

3) dimensions small with respect to earth

Then (being somewhat sloppy in sign convention and assuming m2> m1))

(m2 + m1)* 0.9 g = (m2 - m1) g   [acceleration of entire set of masses is the difference between their gravitational forces]

rearranging terms:

(m2-m1)/(m2+m1) = 0.9

(1-m1/m2)/(1+m1/m2) =0.9

(1-m1/m2) = 0.9 * (1+m1/m2)

0.1= 1.9 * (m1/m2)

m1/m2 = 1/19

Hope this helps.
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