1.
The acceleration of the system is
a=F/(m1+m2) = 58/(15+17)=1.81 m/s².
T1+T2
m1•a=T
m2•a= F-T,
T = m1•a = 15•1.81=27.5 N.
2. μ = 0.09,
T1=T2=T,
m1•a=T-F1(fr)=T- μ•N1=T- μ•m1•g
m2•a=F-T-F2(fr)= F-T- μ•N2= F-T- μ•m2•g
a•(m1+m2) = T- μ•m1•g+ F-T- μ•m2•g =F- μ•g(m1+m2),
a= {F- μ•g(m1+m2)}/ (m1+m2)=
={58 – 0.09•9.8(15+17)}/(15+17) =
=0.93 m/s²
T= m1•a+ μ•m1•g=m1(a+ μ•g) =
=15(0.93+0.09•9.8)=27.18 N
Two boxes of fruit on a frictionless horizontal surface are connected by a light string as in the figures below, where m1 = 15 kg and m2 = 17 kg. A force of 58 N is applied to the 17 kg box.
Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.09.
a) find the acceleration
b) find the tension in the string
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