The coefficient of friction of an inclined plane is 1/√3. If it is inclined at an angle 30⁰ with the horizontal, what will be the downward acceleration of the block placed on the inclined plane?
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The downward parallel force is mg sin30 degree. ( here parallel means parallel to inclined plane)
Upward parallel force is umgcos 30 degree.(u=1/root 3) is coefficient of friction)
Equation of motion: ma=(mg/2) - (1/root 3)mg(root 3/2) or
a=(10/2)-(10/2)=0.
Upward parallel force is umgcos 30 degree.(u=1/root 3) is coefficient of friction)
Equation of motion: ma=(mg/2) - (1/root 3)mg(root 3/2) or
a=(10/2)-(10/2)=0.
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