Resposta:
[tex]\boxed{\theta\approx0.149\ \text{radianos}\approx8.531\º}[/tex]
Explicação:
Supondo que o bloco está estático e analisando o diagrama de forças, percebemos que:
[tex]\boxed{N=P_y}\ \ \text{e}\ \ \boxed{F_{at}=P_x}[/tex]
Assim:
[tex]N=P_y\ \therefore\ N=mg\cos{\theta}\\\\ F_{at}=\mu N\ \therefore\ \boxed{F_{at}=\mu mg\cos{\theta}}[/tex]
[tex]F_{at}=P_x\ \therefore\ \mu mg\cos{\theta}=mg\sin{\theta}\ \therefore\\\\ \mu\cos{\theta}=\sin{\theta}\ \therefore\ \mu=\dfrac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\ \therefore\ \boxed{\boxed{\theta=\arctan{\mu}}}[/tex]
Para [tex]\mu=0.150[/tex]:
[tex]\theta=\arctan{(0.15)}\ \therefore\ \boxed{\theta\approx0.149\ \text{radianos}\approx8.531\º}[/tex]
