Answer:
42.99°
Explanation:
[tex]F_h[/tex] = Kinetic friction force
[tex]F_{\theta}[/tex] = Pulling force at angle [tex]\theta[/tex]
[tex]N_h[/tex] = Weight of the box = 150 N
Kinetic friction force
[tex]F_h=\muN_h[/tex]
Pulling force at angle [tex]\theta[/tex]
[tex]F_{\theta}=\muN_{\theta}[/tex]
N = Pulling force
According to question
[tex]\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N[/tex]
Applying Newton's second law in the vertical direction we get
[tex]N_h-Nsin\theta=N_{\theta}\\\Rightarrow 150-110sin\theta=75\\\Rightarrow \theta=sin^{-1}\frac{75}{110}\\\Rightarrow \theta=42.99\ ^{\circ}[/tex]
The angle is 42.99°