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A 0.305 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= 1.52 NθL=31 degrees 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? Fb= _____Nθb=_____degree

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Answer

given,

[tex]F_L= 1.52\ N[/tex]

[tex]\theta_L= 31^0[/tex]

mass of book = 0.305 Kg

so, from the diagram attached  below

[tex]F_L cos {\theta_L} + F_b sin {\theta_b} = m g[/tex]

[tex]1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8[/tex]

[tex] F_b sin {\theta_b} = 2.989 -1.303[/tex]

[tex] F_b sin {\theta_b} = 1.686[/tex]

computing horizontal component

[tex] F_b cos {\theta_b} = F_L sin {\theta_L}[/tex]

[tex]cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}[/tex]

[tex]cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}[/tex]

[tex]cos {\theta_b} = 0.464[/tex]

θ = 62.35°

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