forceonman=880=Mg+ma=M(9.8+1)
forceonmanconstantspeed=1000=(M+m)g
well, from the Man data, M = 880/(10.8)
put that into the second equation, and solve for m, the son mass.
A man of mass M kg and his son of mass m kg are standing in a lift. When the lift is accelerating upwards with magnitude 1ms-2 the magnitude of the normal contact force exerted on the man by the lift floor is 880N. When the lift is moving with constant speed the combined magnitude of the normal contact forces exerted on the man and the boy by the lift floor is 1000N. Find the values of M and m.
2 answers
AS 880N IS THE FORCE EXERTED BY THE LIFT TO THE MAN
THEREFORE,880-Mg=Ma
880=M(g+a)
880/g+a=M
880/11=80=M
M=80kg
880+N2=1000
N2=120N
N2=m(g+a)
120/11=m
m=11kg
THEREFORE,880-Mg=Ma
880=M(g+a)
880/g+a=M
880/11=80=M
M=80kg
880+N2=1000
N2=120N
N2=m(g+a)
120/11=m
m=11kg
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