vertical problem:
normal force = weight - string up force
= 890 - T sin 30 = 890-.5 T
horizontal problem:
T cos 30 - .2(890-.5T) = (890/9.81) (0.8)
solve for T
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
1 answer