It's Heavy

Classical Mechanics Level 3

In Newtons ( $\text N$ ), how much horizontal force should be supplied to $A$ , so that $A$ moves to the left with an acceleration of $3\text{ m/s}^2$ ?

Give your answer to the nearest integer.

Take gravity to be $9.8\text{ m/s}^2$ .

The answer is 405.

2 solutions

Oliver Garcia
Aug 8, 2014

$For\quad A\\ \sum { { F }_{ y }={ F }_{ na }{ -M }_{ a }g } ={ M }_{ a }{ a }_{ ay }\\ { a }_{ ay }=\quad 0\\ \Longrightarrow \\ \sum { { F }_{ y }=\quad { F }_{ na }{ -M }_{ a }g } =0\\ { F }_{ na }{ =M }_{ a }g\\ \sum { { F }_{ x }= } F-T-F_{ f }={ M }_{ a }a\\ F-T-F_{ f }=\quad { M }_{ a }a\\ T\quad =F-{ M }_{ a }a-F_{ f }\\ \\ For\quad B\\ \sum { { F }_{ x }=T{ -M }_{ b }g } ={ M }_{ b }a\\ T{ =M }_{ b }(g+a)\\ F-{ M }_{ a }a{ -F_{ f }=M }_{ b }(g+a)\\ F_{ f }=\quad \mu { F }_{ na }=\mu M_{ a }g\\ F-{ M }_{ a }a{ -\mu M_{ a }g=M }_{ b }(g+a)\\ F={ M }_{ a }(a{ +\mu g)+M }_{ b }(g+a)\\ F=404.8\quad Newtons$

good image :D

Oliver Garcia - 6 years, 10 months ago

Andi M
Mar 10, 2014

fs - T = ma .a

T - Wb = mb.a

a = Wb -fs / ma + mb

F = ma

m(-a) = Wb + fs - F/ ma + mb

F = m.a + fs + Wb / ma + mb

hmm.. actually i had known the way to solve those problem, but, i don't know why my answer is 404.8 Newton.

Hendra Gusmawan - 6 years, 11 months ago

the answer is 404.8 Newton!!!!, are not 406

Oliver Garcia - 6 years, 10 months ago

@Andi M Do you agree that the answer should be 404.8?

Calvin Lin Staff - 6 years, 10 months ago

The answer is 404.8 newton.

Ronak Agarwal - 6 years, 9 months ago

@Ronak Agarwal – Its 408.8 N

Kushagra Sahni - 6 years, 9 months ago

Much better solution..

bittu thakur - 5 years ago

It's Heavy

The answer is 405.

2 solutions

0 pending reports