Processing Math: Done

From Mechanics

We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector a.

Applying Newton´s Second Law F=ma horisontally,

 : 50cos30−F=maF=50cos30−ma=500.866−2005=43.3−10=33.3 N

We now use the friction condition F=R, however we first need to calculate R.

Note that R is NOT equal to mg here as the string has a component force in the vertical direction! In fact one has,

 : R+50sin30−mg=0R=mg−50sin30=209.81−5021=196.2−25=171.2 N

Using the friction equation gives

33.3=171.2=33.3171.2=0.195