We write the equations of motion for each body in projections on the coordinate axes
на ось x
−Ffr+T=m1a (1)
на ось y
m2g−T=m2a (2)
m1g−N=0 (3)
given that Ffr=μN from the equation (3) we write Ffr=μm1g
add (1) to (2)
m2g−Ffr−T+T=m2a+m1a
or
m2g−μm1g=m2a+m1a
g(m2−μm1)=(m2+m1)a
Then the acceleration is
a=m2+m1g(m2−μm1)=10+49.81(10−0.3⋅4)=6.166m/s2
from (2) we determine the string tension
T=m2g−m2a=m2(g−a)=10(9.81−6.166)=36.44N
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