Please help,thanks in advance Figure 3 shows blocks A & B connected by flexible,
ID: 1712662 • Letter: P
Question
Please help,thanks in advance
Explanation / Answer
a)let tension in chord be T
from free body diagram of block A,we get
300*9.8*(3/5) -T =300a...(i)
From free body diagram of block B,we get
2T-200*9.8*(4/5) = 200a....(ii)
Solving (i) and (ii) we get
T=1029 N
b)since the blocks will be moving,kinetic friction force will be mobilied
Considering free body diagram of block A
300*9.8*(3/5) -T-(0.2*300*9.8*4/5)=300a....(i)1293.6
Considering free body diagram of block B
2T-200*9.8*4/5-(0.3*200*9.8*3/5) =200a....(ii)1920.8
Solving (i) and (ii) we get
T=1043.7
c)friction force on block A=0.2*300*9.8*4/5=470.4 N(upwards along the incline)
Friction force on block B=0.3*200*9.8*3/5=352.8 N(acting downwards along the incline)
d)the blocks will be stationary if for the given values of static friction values, the inclination of the inclines are adjusted such the net force acting on the blocks is zero and static friction is mobilised