A box weighing 890N is pulled along a horizontal surface by means of a string which is at 30o above the horizontal. If the coefficient of kinetic friction is 0.2 and the box is accelerated at 0.8 m/s2, what is the tension in the string?

Answer 1

M*g = 890, M = 890/9.8 = 90.8 kg.

Fn = Mg*cos 0-T*sin30 = 890 - 0.5T.

Fk = u*Fn = 0.2(890-0.5T) = 178-0.1T.

T*Cos30-Fk = M*a.
0.866T-(178-0.1T) = 90.8*0.8,
0.866T-178+0.1T = 72.64,
T = 259.5 N. = Tension in the string.