Venkat
Last Activity: 6 Years ago
ΣFy = mBaB,y
0 = 2T − mBg
T = mBg/2 = 20 kg(9.81 N /kg) /2 = 98.1 N
Summing the tangential forces on mass A,
ΣFt = MAaA,t
0 = T + FF − mAg sin (30◦ )
FF = −T + mAg sin (30◦ ) = 196.2 N
Summing the forces in the normal direction,
ΣFn = MAaA,n
0 = N − mAg cos (30◦ )
N = mAg cos (30◦ ) = 510 N
FF,max = µsN = 127 N
Now set up the force balance equations again, but this time there is acceleration and FF = µkN.
First for block A,
0 = ΣFn − mAaa,n = N − mAg cos (30◦ ) − mA aa,n
N = mAg cos (30◦ ) = 510 N
0 = ΣFt − mAaa,t = T + FF − mAg sin (30◦ ) − mAaa,t
T = −FF + mAg sin (30◦ ) + mAaa,t (1)
Now for block B,
0 = ΣFy − mBaB,y = 2T − mBg − mBaB,y
T = mB 2 (g + aB,y) (2)
Setting equation (1) equal to equation (2) and letting aA,t = −2aB,y,
−FF + mAg sin (30◦ ) + mA (−2aB,y) = mB 2 (g + aB,y)
Solving for aB,y,
mAg sin (30◦ ) − mBg 2 − FF = mBaB,y 2 + 2mAaB,y
aB,y = (mAg sin (30◦ ) – mBg/ 2 − FF )/(mB /2 + 2mA )
aB,y = (mAg sin (30◦ ) − mBg /2 − µkN)/( mB/ 2 + 2mA )
aB,y = (294.4 N − 98.1 N − 101.9 N)/ (10 kg + 120 kg) = 0.726m/ s 2
aA,t = −2aB,y = −1.45 m/ s 2
T = mB 2 (g + aB,y) = 105.4 N
T = −FF + mAg sin (30◦ ) + mAaa,t = 105.4 N
