Question

A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk = 0.53, respectively.
2
29◦
What is the frictional force acting on the 46 kg mass?

Answer 1

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$

where

m = 46 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=29^{\circ}$ is the angle of the incline

- The (static) frictional force, acting upward, of magnitude $F_f$

Since the block is in equilibrium, we can write

$mg sin \theta - F_f = 0$

And substituting, we find the force of friction:

$F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N$

Learn more about frictional force along an inclined plane:

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