If the actual frictional force is equal to 1 lb., then how much effort (force) would be required to drag a 90 lb. box up the plane? a. 10+ lb. b. 10- lb. c.9+ lb. d. 9- lb. could you also explain this? i am having a hard time understanding.

Resolving 90 lb along the incline;

The resultant weight along the incline = 90 Sin ∅ = 90*H/90H = 90*1/9 = 9 lb.

The effort must overcome the resultant weight along the incline as well as the frictional force.
That is,
Fe > 9+1
Fe > 10 lb

Therefore,
Fe = 10+ lb

And the correct answer is A.