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Walter Arfeuille of Belgium lifted a 281.5 kg load off the ground using his teeth. Suppose Arfeuille can hold just three times that mass on a 30.0° slope using the same force. What is the coefficient of static friction between the load and the slope?

Respuesta :

W0lf93
Gravitational, g = 9.81 m/s2 
Theta = 30 degrees, using the same force. 
Let u be the coefficient of static friction 
Force applied F = mg 
Static Friction Force = u3mgcos (theta) 
The gravitational Force = 3mgsin (theta) 
So the force equation = 3mgsin (theta) - (u3mgcos (theta) + mg) = 0 
So 3mgsin (theta) = (u3mgcos (theta) + mg) => 3sin (theta) = (u3cos (theta) + 1)
 => u = (3sin (30) - 1) / 3cos (30) => u = (3(1/2) - 1) / (3 x 0.866)
 => u = (1.5 - 1) / 2.59 = 0.5 / 2.59 = 0.5 / 2.598
 Coefficient of Static Friction u = 0.19.
onyebuchinnaji

The coefficient of static friction between the load and the slope is 0.42.

The given parameters;

  • mass lifted by Walter, m = 281.5 kg
  • slope of the inclined plane, θ = 30⁰

Using the same force to hold three times that mass on the inclined plane, the normal force on the load is calculated as follows;

let the applied force = F = mg

weight of the load = 3mg

[tex]F_n= Fsin(90) + 3mgcos(\theta)= mgsin(90) + 3mgcos(\theta) \\\\F_n = mg + 3mgcos(\theta)[/tex]

Static frictional force on the load;

[tex]F_s= \mu_s F_n \\\\F_s = \mu_s (3mgcos\theta + mg)[/tex]

The parallel force on the load is calculated as follows;

[tex]3mgsin(\theta) - \mu_s (3mgcos\theta + mg) = 0\\\\mg(3sin\theta) - mg\mu_s (3cos\theta +1) =0\\\\u_s = \frac{mg(3sin\theta)}{mg(3cos\theta + 1)} \\\\\mu_s = \frac{3 sin(\theta)}{3 cos(\theta) \ + \ 1} \\\\\mu_s = \frac{3 sin(30)}{3 cos(30) \ + \ 1}\\\\\mu_s = \frac{3(0.5)}{3(0.866) \ + \ 1} \\\\\mu_s = 0.42[/tex]

Thus, the coefficient of static friction between the load and the slope is 0.42.

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