# Motion and Forces

1. Sep 16, 2007

### BuBbLeS01

1. The problem statement, all variables and given/known data
A 3.54 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.6 N at an angle theta = 15.0° above the horizontal, as shown in the Figure. The coefficient of kinetic friction between the block and the floor is 0.08. What is the speed of the block 3.9 s after it starts moving?

2. Relevant equations
Xcomp= 10.6 cos 15
Ycomp= 10.6 sin 15

I don't know where to go from here :(

2. Sep 16, 2007

### Mindscrape

$$\sum_{i=1}^n F_i = ma$$

3. Sep 16, 2007

### BuBbLeS01

so do I have to find the resultant vector or no?

4. Sep 16, 2007

### learningphysics

Did you draw a diagram? Do $$\Sigma\vec{F} = ma$$ in the vertical and horizontal directions...

5. Sep 16, 2007

### BuBbLeS01

I don't know? I am so lost!

6. Sep 16, 2007

### BuBbLeS01

Okay so I have come up with the net force for the x and y directions...

Fxnet = Fx + Wx + Nx = Fx + 0 + Nx = max
Fynet = Fy + Wy + Ny = Fy - Wy + 0 = may

7. Sep 16, 2007

### BuBbLeS01

Is that set up right?

8. Sep 16, 2007

### learningphysics

For the x-direction you wrote:

$$F_x + N_x = ma_x$$

What is N_x ? Does the normal force act in the x - direction? what about the frictional force?

For the y-direction you wrote:

$$F_y - W_y + 0 = ma_y$$

Hmmm... why didn't you put a normal force here?

9. Sep 16, 2007

### BuBbLeS01

Nx doesn't act in the x direction so I put it as 0...
so I have...
Fx/m = ax

For the normal force in the y direction I put 0...
(Fy-Wy)/m = ay

10. Sep 16, 2007

### learningphysics

Hmmm... you're missing forces.

List all the forces acting in the x-direction.

List all the forces acting in the y-direction.

11. Sep 16, 2007

### BuBbLeS01

Forces in the x-direction:

Tension
Normal
Fricton

Forces in the y-direction:

Weight
Normal
Friction
Tension

12. Sep 16, 2007

### learningphysics

There's no normal force in the x - direction. There's no frictional force in the y-direction.

So in the x-direction... you only have the x-component of tension, and the friction... write out the F=ma equation in the x-direction.

In the y-direction... you have the y-component of tension, gravity and the normal force... write out the F=ma equation in the y-direction.

13. Sep 16, 2007

### BuBbLeS01

Oh okay...that was a dumb mistake I made...

Fxnet = Tx + Fx = ma
Fynet = Ty - Wy + Ny = ma

14. Sep 16, 2007

### learningphysics

Cool.. try to substitute the values of tension, friction, weight etc... into your equations...

15. Sep 16, 2007

### BuBbLeS01

I got the friction and weight but I m drawing a blank on how to calculate the tension?

16. Sep 16, 2007

### learningphysics

Actually you have the components of tension in your first post.

17. Sep 16, 2007

### BuBbLeS01

okay so the fx = 10.6 cos 15 is actually the tension force not the friction force?

18. Sep 16, 2007

### learningphysics

yes, 10.6 is the force the cord exerts... ie tension... so 10.6cos15 is the horizontal component... 10.6sin15 is the vertical component.

19. Sep 16, 2007

### BuBbLeS01

then how do I calculate the friction force. I know I have to use the Kinetic friction force of 0.08 but isn't there something else I have to do to it?

20. Sep 16, 2007

### learningphysics

Frictional force = normal force * uk

uk is the coefficient of kinetic friction = 0.08