Question

Two masses are connected by a string over the edge of a table by a frictionless pulley. The horizontal table holds the second mass m2, and the first mass m1 is suspended off of the pulley. Assume the table is frictionless unless specified otherwise.

For each part of this problem: Draw free body diagrams of both objects and write down equations for Newtons second law for both masses.

(a) [10 points] Calculate the acceleration if m1 = 5 kg and m2 = 0 kg

(b) [10 points] Calculate the acceleration if m1 = 0 kg and m2 = 5 kg

(c) [10 points] Calculate the acceleration if m1 = 5 kg and m2 = 3 kg

(d) [10 points] Assume the table is rough (include friction) and m1=5kg and m2=5kg. What is the minimum coefficient of friction required to keep the block at rest on the top of the table.

Answer 1

T fat T ta mag ca. m,= skg and ma m2 = оку assume block mi is moving down is moving sight. and ma for mi Newton's 2nd law equation mig -T = mia - and for m2 m2 ) T= m29 By adding 0 4 0 we get & mig la = = mi tm a) a mig (A mitmz Now mi= sky and m2=0 kg : a (a=9.8m/s2 (5x9,8 $+0 and m 2 = mia ong skg QE ох 0 sto a=0 m/s2

mi- his and maz 3 kg ma bu matm = 6.125 m/s2 5 x 9.8 (5+3) 6.125 m/s² Р) Now let'ts' be static friction acting on it mia sig, m2 = 5kg so to kap rest (ã=o) for block 'ma' Newton 2nd law egr T fs m 29 0 = sf - 1 and for block m1 => mg =T N= bru (T= sxg T = TE T- MsXN MSX meg A S9= Msx sx M$=1