5) applies, but with theaddition of an “uphill” tension force T (as in Fig. 5-18

ID: 1762924 • Letter: 5

Question

5) applies, but with theaddition of an “uphill” tension forceT (as in Fig. 5-18(b)) and

withfsreplaced with fk,incline (tobe as general as possible, we are treating the inclineas

having a coefficient ofkinetic friction ). If wechoose “downhill” positive, then

Newton’s lawgives

mA g sin – fA– T =mAa

for blockA (where =30º). For block B we chooseleftward as the positive direction and

writeT – fB= mBa. Now

fA =k,incline FNA= mAg cos

using Eq. 6-12 applies toblock A, and

fB =k FNB= kmBg.

In this particularproblem, we are asked to set = 0, and the resulting equations can be

straightforwardly solvedfor the tension: T= 13 N.

(b) Similarly, finding thevalue of ais straightforward:

a = g(mAsin – kmB)/(mA+ mB) =1.6m/s2.

Thanks for the help yet again......

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