Calculating Resultant of Forces on 6.2KG Box in Lift

[DeanBH]

A box of mass 6.2KG stands on the floor of a lift which is moving with an upward acceleration of 0.2ms^-2. A horizontal force of magnitude 16N acts on the box. There is no horizontal movement of the box.

Find the magnitude and direction of the resultant of the forces, normal and frictional, exerted by the floor on the box.

i had a url but it won't let me say them without 15 posts so i can't show you what I've done already.

 

[sokratesla]

# If there is no motion in a direction, then the sum of the forces in that direction is zero.
# Here there is no motion in the horizontal direction. The other for in the horizontal direction is the friction. From this info you can find the friction.
# The force which makes the box go up is the normal force. Normal force both accelerates the box and hold its weight. For z direction you should write the sum of normal force and the weight, but this sum is this time not equal to zero.
# I hope these will help you.

 

[_Mayday_]

DeanBH, I think you still need to show some type of attempt.