A stubborn, 120 kg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around the mule and pulls with his maximum force of 800 N. The coefficients of friction between the mule and the ground are ms = 0.8 and mk = 0.5. Is the farmer able to move the mule?

Respuesta :

Answer: No , he is definitely not

Explanation:

The mule is sitting down, the forces interacting with it vertically are its weight (120kg*9.8m/s^2) which is equal to the normal force  N = 120kg*9.8m/s^2 = 1176N. Horizontally we have the 800N exerted by the farmer minus ms*N (static friction force), according to the theory, in order to move the mule the farmer must satisfy [tex]m_{s} = F_{smax}/N[/tex] which means 0.8*1176N = F[tex]F_{smax}[/tex] = 940 N which is not what the farmer is pulling, therefore, the mule will not move an inch

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