A force F pulls a block of mass 60 kg at a c | vedclass

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(B) Given: mass $m = 60 \ kg$,angle $	heta = 37^{\circ}$,coefficient of friction $\mu = 1/3$,$g = 10 \ m/s^2$.Since the block moves at a constant spe

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$200$

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(B) Given: mass $m = 60 \ kg$,angle $	heta = 37^{\circ}$,coefficient of friction $\mu = 1/3$,$g = 10 \ m/s^2$.Since the block moves at a constant speed,the net force is zero $(a = 0)$.Resolving the force $F$ into horizontal and vertical components:Horizontal direction: $F \cos 37^{\circ} = \mu N$  --- $(1)$Vertical direction: $N + F \sin 37^{\circ} = mg$  --- $(2)$From $(2)$,$N = mg - F \sin 37^{\circ}$.Substitute $N$ into $(1)$:$F \cos 37^{\circ} = \mu (mg - F \sin 37^{\circ})$$F \cos 37^{\circ} = \mu mg - \mu F \sin 37^{\circ}$$F (\cos 37^{\circ} + \mu \sin 37^{\circ}) = \mu mg$$F = \frac{\mu mg}{\cos 37^{\circ} + \mu \sin 37^{\circ}}$Using $\cos 37^{\circ} = 4/5$ and $\sin 37^{\circ} = 3/5$:$F = \frac{(1/3) 	imes 60 	imes 10}{(4/5) + (1/3) 	imes (3/5)} = \frac{200}{(4/5) + (1/5)} = \frac{200}{1} = 200 \ N$.

$A$ force $F$ pulls a block of mass $60 \ kg$ at a constant speed across the floor. The coefficient of friction is $\mu = 1/3$. What is the magnitude of the force $F$ (in $N$)?

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