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

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

 

[rwisz]

Hello, it seems like you are on the right track with your attempted solution. Let's break down the problem and go through each step together.

First, we know that the block is moving at constant velocity, which means that the net force acting on it must be zero. This is because according to Newton's first law, an object will continue to move at a constant velocity unless acted upon by a net force.

Next, we can draw a free body diagram to represent the forces acting on the block. We have the applied force F, the weight of the block mg, and the frictional force u_kmg. The weight of the block is acting downwards, while the applied force and frictional force are acting horizontally.

Using Newton's second law, we can set up two equations: one for the y-axis and one for the x-axis. For the y-axis, we have F_y - mg = ma = 0. Since the block is not moving up or down, the acceleration in the y-direction is zero. This means that F_y must equal mg, or 400 N.

For the x-axis, we have F_x - u_kmg = ma = 0. Since the block is moving at constant velocity, the acceleration in the x-direction is also zero. This means that F_x must equal u_kmg, or (0.4)(400 N) = 160 N.

But wait, the given answer is 150 N! What happened to the extra 10 N? This is where we need to take into account the direction of the forces. The applied force F is acting in the same direction as the motion of the block, while the frictional force is acting in the opposite direction. This means that the net force in the x-direction is actually F_x - u_kmg = 160 N - 40 N = 120 N. And since we know that the net force must be zero, we can set up the equation F_x - u_kmg = 0 and solve for F_x to get F_x = u_kmg = 0.4(400 N) = 160 N. And finally, we can use the Pythagorean theorem to find the magnitude of the applied force: F = sqrt(F_x^2 + F_y^2) = sqrt((160 N)^2 + (400 N)^2) = 150 N.

I hope this helps clear up any confusion you had. Remember to always pay