Solution 10.7

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,

 : 40cos30−F=ma ration is represented by the vector a.

We need to obtain F.

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+40sin30−mg=0F=mg−40sin30=39.81−4021=29.43−20=9.43 N

Using the friction equation gives

F=RF=0.59.43=4.71 N

Substituting in the above equation of motion (Newton´s Second Law) gives,

3a=400.866−4.71=29.93a=9.98 ms−2