A worker provides an acceleration of a = 0,31 m/s2 by pushing two containers of masses m1 = 71 kg and m2 = 31 kg next to each other on a horizontal surface. If the coefficient of friction between the containers and the surface is 0.15 calculate the worker's applied force(in nunit of N) on the container. (Use g=10 N/kg)
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Ответ:
Объяснение:
a = 0,31 м/с^2
m1 = 71 кг
m2 = 31 кг
μ = 0,15
g=10 Н/кг
F = ?
сила под действием которой движутся ящики
F1 = a* (m1 + m2) = 0.31* (71+ 31) = 31.62 H
Fтр = INI*μ = m*g*μ в общем виде
сила трения для обоих ящиков
Fтр =g*μ * (m1 + m2) = 10*0.15* (71+31) = 153 H
тогда сила приложенная рабочим
F = Fтр + F1 = 153+31.62 = 184.62 H