A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling force can point horizontally, or it can point above the horizontal at an angle θ. When the pulling force points horizontally, the kinetic frictional force acting on the box is twice as large as when the pulling force points at the angle θ. Find θ.

Respuesta :

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°

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