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
[tex]mg sin \theta[/tex]
where
m = 46 kg is the mass
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\theta=29^{\circ}[/tex] is the angle of the incline
- The (static) frictional force, acting upward, of magnitude [tex]F_f[/tex]
Since the block is in equilibrium, we can write
[tex]mg sin \theta - F_f = 0[/tex]
And substituting, we find the force of friction:
[tex]F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N[/tex]
Learn more about frictional force along an inclined plane:
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