Question

A 60 N tool box is dragged horizontally with constant speed by a rope that forms a 35° angle with the floor. The tension measured is 40 N. Find the magnitude of the friction force and the normal force. What is the kinetic friction coefficient?

Answer 1

The magnitude value of the frictional force of the normal is

Fr = 32.76 N; Fn = 37.06 N and the coefficient of kinetic friction is

u = 0.88

Explanation:

We carry out sum of forces, in both axes:

∑Fy: 0

  40Sen35 ° + Fn - w = 0

 40Sen35 ° - 60N = -Fn

  Fn = 37.06 N

∑Fx: 0

  40NCos35 ° - Fr = 0

  Fr = 32.76 N

Friction force

Fr = uFn

u = Fr / Fn

u = 32.76N / 37.06N

u = 0.88

El valor de magnitud de la fuerza de friccion de la normal es

Fr = 32.76 N ; Fn  = 37.06 N y el coeficiente de roce cinetico es

u =0.88

Explicación:

Realizamos sumatoria de fuerzas,en ambos ejes:

∑Fy : 0

  40Sen35° + Fn - w = 0

 40Sen35° - 60N = -Fn

 Fn  = 37.06 N

∑Fx : 0

  40NCos35° - Fr = 0

  Fr = 32.76 N

Fuerza de friccion

Fr = uFn

u = Fr/Fn

u = 32.76N/37.06N

u =0.88