Newton's laws - studying for a test

1. Sep 19, 2011

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.

2. Sep 19, 2011

PhanthomJay

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

3. Sep 19, 2011

computerex

oops. Sorry, here is a picture:

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

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.

Last edited by a moderator: May 5, 2017
4. Sep 19, 2011

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

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