A 0.135 kg book rests at an angle against one side of a bookshelf. The magnitude
ID: 1418512 • Letter: A
Question
A 0.135 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the book by the left side of the bookshelf are given by:
FL = 0.767 N ThetaL = 59degrees
What must the magnitude and direction of the total force exerted on the book by the bottom of the bookshelf be in order for the book to remain in this position?
A 0.135 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the book by the left side of the bookshelf are given by: FL-0.767 N What must the magnitude and direction of the total force exerted on the book by the bottom of the bookshelf ' = 59.0 be in order for the book to remain in this position? Number NumberExplanation / Answer
F(L)x = 0.767 N * sin 59 = 0.657 N (right)
F(L)y = 0.767 * cos 59 = 0.395 N (up)
Net Force x = 0.767 N * sin 59 - F(B) cos (B) = 0
Net Force y = 0.767 N * cos 59 + F(B) sin (B) - 0.135 kg*9.8 = 0
F(B)x = F(B) cos (B) = 0.657 N (left)
F(B)y = F(B) sin (B) = 0.135 kg*9.8 - 0.767 N * cos 59
F(B)y = 0.928 N (up)
F(B) = sqrt ( F(B)x^2 + F(B)y^2 ) = 1.137 N
(B) = arctan (F(B)y / F(B)x) = arctan (0.928 N / 0.657N)
(B) = 54.7 degree
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