Force on an object being dragged

[jj8890]

Homework Statement

You drag a suitcase of mass 9.2 kg with a force of F at an angle 52.1 degrees with respect to the horizontal along a surface with kinetic coefficient of friction 0.54. The acceleration of gravity is 9.8 m/s^2.
If the suitcase is moving with constant velocity 1.65 m/s, what is F ?

Homework Equations

F=ma

The Attempt at a Solution

F=ma

a = 0

FcosΘ - Friction = (9.2)(0)
F - µR = 0

F - (0.54)((9.2)(9.8)-mgsinΘ)) = 0
F- (0.54)(90.16 - 9.2(9.8)sin52.1) = 0
F - 0.54(19.016) = 0
F = 10.268 N

but this is not right..I think that I am doing something wrong...Any help would be great. Thanks

 

[physixguru]

this should be enough...

If the object moves with constant velocity then NET force on it is zero.
REMEMBER THE KEYWORD "NET" HERE...

Fcos theta - friction= zero

>fcos theta = frictional force.
>f cos theta=48.68
>f=48.68/cos 52.1
>f=48.68/.614
>f=79 N

 

[kamerling]

this should be enough...

If the object moves with constant velocity then NET force on it is zero.
REMEMBER THE KEYWORD "NET" HERE...

Fcos theta - friction= zero

>fcos theta = frictional force.
>f cos theta=48.68
>f=48.68/cos 52.1
>f=48.68/.614
>f=79 N

The frictional force depends on F and theta. If you pull upwards the suitcase, the force that the suitcase exerts on the ground will be smaller, and therefore the friction will decrease.
with theta = 60 your answer would have F > mg and the suitcase flying up in the air.

The force with which you pull upward on the suitcase is F sin(theta), so the frictional force is 0.54 (mg - F sin(theta))