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(B) The angle of friction is $\beta$,so the coefficient of friction is $\mu = 	an \beta$.When a pushing force $F$ is applied at an angle $\alpha$ wit

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$\frac{m g \sin \beta}{\cos (\alpha+\beta)}$

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(B) The angle of friction is $\beta$,so the coefficient of friction is $\mu = 	an \beta$.When a pushing force $F$ is applied at an angle $\alpha$ with the horizontal,the normal force $N$ is given by balancing the vertical forces:$N = m g + F \sin \alpha$To just move the block,the horizontal component of the force must equal the limiting friction:$F \cos \alpha = \mu N$Substituting $\mu = 	an \beta$ and $N = m g + F \sin \alpha$:$F \cos \alpha = 	an \beta (m g + F \sin \alpha)$$F \cos \alpha = \frac{\sin \beta}{\cos \beta} (m g + F \sin \alpha)$$F \cos \alpha \cos \beta = m g \sin \beta + F \sin \alpha \sin \beta$$F (\cos \alpha \cos \beta - \sin \alpha \sin \beta) = m g \sin \beta$Using the trigonometric identity $\cos (A + B) = \cos A \cos B - \sin A \sin B$:$F \cos (\alpha + \beta) = m g \sin \beta$$F = \frac{m g \sin \beta}{\cos (\alpha + \beta)}$

If a pushing force making an angle $\alpha$ with the horizontal is applied on a block of mass $m$ placed on a horizontal table and the angle of friction is $\beta$,then the minimum magnitude of the force required to move the block is

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