Two blocks of masses m1=5 and m2=6 kg are on either side of the wedge shown below. Assume the surface and the pulley are frictionless. Find
A)the direction this system will accelerate
B) the acceleration of m1 and m2
C) the tension in the rope
did I do this right? how do I find the direction
M1
Fx=T-m1gsin30= m1a=plug in tension of the rope= 36.5-5kg(9.8)sin30=5a a=2.4
Fy:Fn1-m1gcos30=m1a=0
M2
Fx=m2gsin60-T=plug in tension of the rope=6kg(9.8)sin60-36.5=14.4222937424992=6a=2.4
Fy=Fn2-m2gcos60=0
T-m1gsin30=m1a
-t+M2gsin60=m2a
A=(T-m1gsin30)/m1
Combine eq
-t+M2gsin60=m2((T-m1gsin30)/m1))
-t+6kg(9.8)sin60=6kg((T-5*9.8sin30)/5)
-t+50.9= 6((T-24.5)/5)
-t+50.9 = 6/5t -29.4
80.3=2.2t
T=36.5N
1 answer
yes