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A tired worker pushes with a horizontal force of 440 N on a 170 kg crate initially resting on a thick pile carpet. The coefficients of static and kinetic friction are 0.5 and 0.2, respectively. The acceleration of gravity is 9.81 m/s^2 . Find the frictional force exerted by the carpet on the crate. Answer in units of N.

Respuesta :

Answer:

440 N

Explanation:

Force applied, F = 440 N

mass of crate, m = 170 kg

μs = 0.5, μk = 0.2

g = 9.81 m/s^2

The normal reaction acting on the crate, N = m g = 170 x 9.81 = 1667.7 N

The maximum value of static friction force acting on the crate

[tex]f_{s}=\mu _{s}N=0.5 \times 1667.7 = 833.85 N[/tex]

The maximum value of static friction force is more than the applied force so the crate does not move and teh applied force becomes friction force.

thus, the friction force acting on the crate is 440 N.

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