What is the acceleration of a bar sliding along an inclined plane if its angle of inclination is 30 ‘degrees and the coefficient of friction is 0.15?
Expand the acceleration g into two axes: OX and OY
g (y) = g * cos a
g (x) = g * sin a
Friction force (Ftr)
Ftr = mu * m * g * cos a
F tightness by OX = F1 = mg * sin a
According to Newton’s Laws:
F1-Ftr = m * a
g * sin a – mu * g * cos a = a (acceleration)
Answer: g * sin a – mu * g * cos a = a (acceleration)
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