The body

A body slides down an inclined plane at an angle of α = π/4 = 45° to the horizontal, with friction causing a deceleration of a = 2.4 m/s². At what angle β must the plane be inclined so that, after a small push, the body slides at a constant speed?

Final Answer:

B =  33.1853 °

Step-by-step explanation:

$$\begin{gathered} \alpha =\frac{\pi }{4}=\frac{3.1416}{4}\doteq 0.7854\text{rad} \\ a=2.4{\text{m/s}}^2 \\ g=9.81{\text{m/s}}^2 \\ G=mg \\ {F}_1=ma=m\cdot g\cdot \sin \alpha -m\cdot g\cdot k\cdot \cos \alpha \\ ma=m\cdot g\cdot \sin \alpha -m\cdot g\cdot k\cdot \cos \alpha \\ a=g\cdot \sin \alpha -g\cdot k\cdot \cos \alpha \\ k=(g\cdot \sin \alpha -a)/(g\cdot \cos \alpha )=(9.81\cdot \sin 0.7854-2.4)/(9.81\cdot \cos 0.7854)\doteq 0.654 \\ {a}_2=0{\text{m/s}}^2 \\ g\cdot \sin \beta -g\cdot k\cdot \cos \beta =0 \\ \sin \beta -k\cdot \cos \beta =0 \\ \sin \beta /\cos \beta -k=0 \\ \tan \beta -k=0 \\ \beta =\arctan k=\arctan 0.654\doteq 0.5792\text{rad} \\ B=\beta \to \text{°}=\beta \cdot \frac{180}{\pi }\text{°}=0.5792\cdot \frac{180}{\pi }\text{°}=33.185\text{°}=33.1853^{\circ }=33°11^{\prime }7" \end{gathered}$$

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