What is the force?

How much force (N) are given to A, so system moved to the left with acceleration 3 m/s^2 ?

Please first obtain a general equation, then replaced one decimal

This problem is not original, the original problem has a bug The original problem is ---> Original Problem


The answer is 404.8.

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1 solution

Oliver Garcia
Aug 12, 2014

F o r A F y = F n a M a g = M a a a y a a y = 0 F y = F n a M a g = 0 F n a = M a g F x = F T F f = M a a F T F f = M a a T = F M a a F f F o r B F x = T M b g = M b a T = M b ( g + a ) F M a a F f = M b ( g + a ) F f = μ F n a = μ M a g F M a a μ M a g = M b ( g + a ) F = M a ( a + μ g ) + M b ( g + a ) F = 404.8 N e w t o n s For\quad A\\ \sum { { F }_{ y }={ F }_{ na }{ -M }_{ a }g } ={ M }_{ a }{ a }_{ ay }\\ { a }_{ ay }=\quad 0\\ \Longrightarrow \\ \sum { { F }_{ y }=\quad { F }_{ na }{ -M }_{ a }g } =0\\ { F }_{ na }{ =M }_{ a }g\\ \sum { { F }_{ x }= } F-T-F_{ f }={ M }_{ a }a\\ F-T-F_{ f }=\quad { M }_{ a }a\\ T\quad =F-{ M }_{ a }a-F_{ f }\\ \\ For\quad B\\ \sum { { F }_{ x }=T{ -M }_{ b }g } ={ M }_{ b }a\\ T{ =M }_{ b }(g+a)\\ F-{ M }_{ a }a{ -F_{ f }=M }_{ b }(g+a)\\ F_{ f }=\quad \mu { F }_{ na }=\mu M_{ a }g\\ F-{ M }_{ a }a{ -\mu M_{ a }g=M }_{ b }(g+a)\\ F={ M }_{ a }(a{ +\mu g)+M }_{ b }(g+a)\\ F=404.8\quad Newtons

The correct answer is 408.8 N

Kushagra Sahni - 6 years, 9 months ago

please check your answer, the correct answer is 404.8 N

Oliver Garcia - 6 years, 9 months ago

Did it the same way.

Kushagra Sahni - 6 years, 9 months ago

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