Newton's laws - studying for a test

[computerex]

Hello guys. I have a test in a couple of days, so I am hoping to get some practice. I'll post any questions I have here, I would appreciate if you guys can help me clear up any confusions. 1. A 400-N block is dragged along a horizontal surface by an applied force F as shown. The coefficient of kinetic friction is uk = 0.4 and the block moves at constant velocity. The magnitude of F is:

attempted solution:

Since the block moves with constant velocity: $$\sum F = 0$$
Since
$$\sum F = ma$$
$$F_y - mg = ma = 0$$
$$F_x - u_kmg = ma = 0$$

The correct answer is 150 N, but obviously the components above are far greater because mg = 400 N. I must be missing something conceptually simple... Please help.

 

[PhanthomJay]

What is the direction of the applied force "as shown"? What answer do you get?

 

[computerex]

What is the direction of the applied force "as shown"? What answer do you get?

oops. Sorry, here is a picture:

[PLAIN]http://dl.dropbox.com/u/42149615/phq.jpg

Solving the simple equations I came up with above, I would get the components of the force as:

Fy = mg
Fx = Ukmg

The magnitude of F will then of course be far greater then 150 N.

 

[issacnewton]

draw a free body diagram. in the vertical direction, there are three forces acting. normal reaction by the floor , (3/5)F , both of which are upwards and mg , which is downwards.
since there is no net force in the vertical direction,

$$N+\frac{3}{5}F=mg$$

$$\therefore N=mg-\frac{3}{5}F$$

so your force of friction which is opposing the applied force is

$$F_{fr}=\mu_k N=\mu_k (mg-\frac{3}{5}F)$$