Asked by melad
A box sits on a horizontal wooden board. The coefficient of static friction between the box and the board is 0.41. You grab one end of the board and lift it up, keeping the other end of the board on the ground. What is the angle between the board and the horizontal direction when the box begins to slide down the board?
Answers
Answered by
Anonymous
njnj
Answered by
Anonymous
22.29
22deg
22deg
Answered by
henry2,
Mg = Force of box = Normal force,Fn.
Fp = Mg*sin A. = Force parallel with incline.
Fs = u*Fn = 0.41Fn = 0.41Mg = Force of static friction.
Fp-Fs = M*a
Mg*sin A-0.41Mg = M*0 = 0.
Divide both sides by Mg:
sin A-0.41 = 0
A = 24.2 Degrees.
Fp = Mg*sin A. = Force parallel with incline.
Fs = u*Fn = 0.41Fn = 0.41Mg = Force of static friction.
Fp-Fs = M*a
Mg*sin A-0.41Mg = M*0 = 0.
Divide both sides by Mg:
sin A-0.41 = 0
A = 24.2 Degrees.
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