# Question about forces, velocities, accelerations, and slopes

1. Oct 20, 2009

### WahooMan

1. The problem statement, all variables and given/known data

An 8.0kg box slides a distance of 4.0m down a plane that is inclined at 30.0 degrees with the horizontal. The initial speed of the box is 2.5m/s, and the coefficient of kinetic friction between the box and the plane is 0.24. Calculate the final speed of the box.

2. Relevant equations

Fg=mg
Fn=mgcos(theta)? This is what my question is about
Ff=mu(Fn)

3. The attempt at a solution

I have solved for Fg, but my question is how do I know when to use sine and when to use cosine in the equation to find Fn? I have noticed sometimes I get the correct answer with one and sometimes I get the correct answer with the other. How do I know when to use which?

2. Oct 20, 2009

### rl.bhat

Component of Fg which is perpendicular to the surface contributes to the normal reaction.
Component of Fg which is parallel to the surface contributes to the frictional force. Free body diagram will help you to decide which component to be used.

3. Oct 20, 2009

### WahooMan

I don't understand. How do I know when to use sine and when to use cosine?

4. Oct 20, 2009

### rl.bhat

Find Fgsinθ and Fgcosθ. Νow see the directions of these components. And decide for what purpose you have to use them.

5. Oct 20, 2009

### WahooMan

Fgsin(theta) = 39.2
Fgcos(theta) = 67.9

sin=y, cos=x?

So if I use sine to find my y component, Fn=39.2N Is this correct?

Last edited: Oct 20, 2009
6. Oct 20, 2009

### rl.bhat

What is the frictional force?
In which direction it acts?
Which component pushes the box along the slope?

7. Oct 20, 2009

### WahooMan

I don't know the frictional force because I don't know how to calculate Fn. I do know that the friction force acts in the negative x direction and the x component pushes the box along the slope. But without knowing how to calculate Fn I can't finish the problem. How do I know when to use sine and when to use cosine?

8. Oct 20, 2009

### rl.bhat

Resolve the components of weight mg along the inclined plane and perpendicular to inclined plane. Perpendicular component contributes to the normal reaction Fn. Horizontal component pushes the box down.

9. Oct 20, 2009

### WahooMan

So Fn = 39.2N perpendicular to slope
Force pushing the box down the slope = 67.9N parallel to slope
Force of friction = Fn(mu) = 39.2(0.24) = 9.408N
Force of box sliding down the slope = 67.9N - 9.408N = 58.492N

Is all of this correct?

10. Oct 20, 2009

### WahooMan

Please help me, I have a lot of studying to do and I've been trying to figure this problem out for more than two hours now. I just need to know in what situations I use sine and when to use cosine, not just in this problem but in general.

11. Oct 20, 2009

### WahooMan

Ok, I've got

Fg = mg = (8)(-9.80) = -78.4N
Fn = mgcos(theta) = 78.4cos30 = 67.9N
Ff = mu(Fn) = 0.24(67.9N) = -16.3N
Fa = mgsin(theta) = 78.4sin30 = 39.2N

Sum of the forces = ma
(39.2+67.9-78.4-16.3) = 12.4 = 8a
a = 1.55 m/s^2

vf^2 = vi^2 + 2a(xf-xi)
vf^2 = 0 + 2(1.55)(4) = 12.4
vf = 3.5 m/s

But my answer key says I should be getting 5.4 m/s What am I doing wrong?

12. Oct 20, 2009

### rl.bhat

While finding out sum of forces, you have to consider only Fa and Ff. Fgcosθ is balanced by normal reaction Fn. So recalculate a.

13. Oct 20, 2009

### WahooMan

Sum of the forces = ma
(39.2-16.3) = 22.9 = 8a
a = 2.86 m/s^2

vf^2 = vi^2 + 2a(xf-xi)
vf^2 = 0 + 2(2.86)(4) = 22.88
vf = 4.78 m/s

I'm getting closer, but I need 5.4 m/s

14. Oct 20, 2009

### rl.bhat

Initial velocity is not equal to zero. Read the problem again.

15. Oct 20, 2009

### WahooMan

Wow.. I got it this time.. I hope I don't make such careless mistakes tomorrow on the test..

So I always use cosine when finding Fn? Would I ever use sine?

16. Oct 21, 2009

### rl.bhat

If you pull of push a block with a force F making an angle θ to the horizontal, then Fsinθ contributes to Fn.