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(B) Given: Mass $m = 4 \ kg$, coefficient of static friction $\mu_s = 0.5$, and frictional force $f = 14.14 \ N = 10\sqrt{2} \ N$.Since the block is a

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$45$

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(B) Given: Mass $m = 4 \ kg$, coefficient of static friction $\mu_s = 0.5$, and frictional force $f = 14.14 \ N = 10\sqrt{2} \ N$.Since the block is at rest on the inclined plane, the frictional force balances the component of gravitational force acting down the plane.$f = mg \sin 	heta$$10\sqrt{2} = 4 	imes 10 	imes \sin 	heta$$10\sqrt{2} = 40 \sin 	heta$$\sin 	heta = \frac{10\sqrt{2}}{40} = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}} \approx 0.3535$.However, if we assume the frictional force given is the limiting friction $(f_L = \mu_s N)$:Normal reaction $N = mg \cos 	heta = 40 \cos 	heta$.$f_L = \mu_s N = 0.5 	imes 40 \cos 	heta = 20 \cos 	heta$.Equating $f_L = 10\sqrt{2}$:$20 \cos 	heta = 10\sqrt{2}$$\cos 	heta = \frac{10\sqrt{2}}{20} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Therefore, $	heta = 45^{\circ}$.

$A$ block of mass $4 \ kg$ is at rest on a rough inclined plane making an angle of $\theta$ with the horizontal. The coefficient of static friction between the block and the plane is $0.5$ and the frictional force on the block is $14.14 \ N$. Find the value of $\theta$. (in $^{\circ}$)

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