A body weighing 3 kg lying on a horizontal surface is acted upon by a horizontal force whose modulus is F = 5 N.

| August 8, 2021 | education

A body weighing 3 kg lying on a horizontal surface is acted upon by a horizontal force whose modulus is F = 5 N. If the coefficient of friction between the body and the surface is 0.20, then the modulus of the friction force acting on the body is equal.

m = 3 kg.

g = 10 m / s2.

F = 5 N.

μ = 0.2.

Ftr -?

Let us find the value of the sliding friction force Ftrsk by the formula: Ftrsk = μ * N, where μ is the friction coefficient, N is the surface reaction force.

Since the body does not move vertically, the reaction force of the support N is compensated by the force of gravity m * g: N = m * g.

The formula for the sliding friction force will take the form: Ftrsk = μ * m * g.

Ftrsk = 0.2 * 3 kg * 10 m / s2 = 6 N.

Since the value of the horizontal force F = 5 N is less than the sliding friction force Ftrk = 6 N, the body will be at rest and the static friction force Ftr = 5 N will act on it.

Answer: the body is affected by the static friction force Ftr = 5 N.



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