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                        i need complete solution for  question..
 
     which i am sending throuh image..kindly consider my request... thank you..

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Grade:11

2 Answers

vishal
22 Points
5 years ago
first find out the tension in the upper string which is connected to block A if the tension induced in the string is more than the limiting friction(static fric)thn there will be relative motion between the block and the plane or else it will remain at rest.the accln of the upper block will nt be the same as that of the lower block....if u take the acc of lower block as a then that of the upper block will be 2a.....once the blocks are in motion consider kinetic friction...hope it helps
Venkat
273 Points
5 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

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