Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :
[tex]mg\ sin\ \theta = 120\times 10 \times sin\ 47^\circ{}=877.62 \ N[/tex]
Horizontal component of force(Normal reaction) :
[tex]mg\ cos\ \theta = 120\times 10 \times cos\ 47^\circ{}=818.40 \ N[/tex]
Since, box is on the verge of slipping :
[tex]mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07[/tex]
Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
