# Homework Help: Simple incline problem has me confused

1. Dec 22, 2014

### keithcuda

1. The problem statement, all variables and given/known data
1. An inclined plane is used to raise a 100 kg box to a height of 5 m. How long must the inclined plane be if a force of 250 N is used to push the box up the inclined plane?

2. Relevant equations
i have search and searched all over google to try to figure this out.. I am not sure if I'm over thinking it or just confused myself even more. I found one equation that I think is correct, but as i was looking at other examples I tried to use this formula and it never worked out like I thought it should. Also, I am always finding examples that have an angle of inclination.. But the formula I think should work is.

$$\frac{f}{w}=\frac{h}{l}$$

3. The attempt at a solution
$$\frac{250}{100}=\frac{5}{l}$$
l=2???

cheers
keith

2. Dec 22, 2014

### Staff: Mentor

Hi keithcuda. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

You won't be able to avoid involving either sinθ or cosθ in your calculations.
Can you attach a pic showing the relevant diagram you have drawn for this problem?

Last edited by a moderator: May 7, 2017
3. Dec 22, 2014

### Jazz

You could draw a free body diagram to see what forces are opposing the motion of the box on an incline. For there to be an upward motion, that or those force(s) must be less than the upward force.

4. Dec 22, 2014

### keithcuda

i can try to draw pic and post it...

That is where i am confused.. the question isn't giving me an angle of the incline.. so how do i include sinθ or cosθ?

5. Dec 22, 2014

### Jazz

Maybe one of the opposing forces is related to the angle $\theta$ of the incline.

6. Dec 22, 2014

### Staff: Mentor

θ
So call the slope θ, or x or something.

7. Dec 22, 2014

### keithcuda

#### Attached Files:

• ###### ?#3.jpg
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Last edited by a moderator: May 7, 2017
8. Dec 22, 2014

### Staff: Mentor

OK. Next show a vector representing the force of gravity on that box. What magnitude would this force have? And its units?

Have you given the slope a name, i.e., a letter?

9. Dec 22, 2014

### keithcuda

OK, so
Ok, so I did some more googling...
I found this, Fp=w(h/l)
Fp=250N
w=mg=100*9.8
h=5
l=slope
250=980(5/l)
l=19.6m ?

10. Dec 22, 2014

### keithcuda

let me know what I am doing wrong or missing and as soon as i leave work and get home I'll work on it some more

cheers

11. Dec 22, 2014

### Staff: Mentor

Once you have answered the questions in post #8 ....

gravitational force on the box can be resolved into an equivalent pair of forces at right-angles to each other: one force acting parallel to the plane, and the second force acting perpendicular to the plane

Forces are vectors, so to add them you must add them as vectors. Draw this "triangle of forces" to show these two perpendicular forces summing to equal the gravitational force on the box. Use trigonometry to calculate the magnitudes of the unknown sides of this triangle.

12. Dec 22, 2014

### keithcuda

Alright, I believe I know what you are talking about... But, I'm just getting home from my power plant after working a 17 hour shift. Once I get some sleep, I'll work on post #8 and #11.

13. Dec 23, 2014

### Staff: Mentor

If you need help with the vector diagram, search on google images for "block on inclined plane". You'll get hundreds. :w

http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg [Broken]

Last edited by a moderator: May 7, 2017
14. Dec 23, 2014

### keithcuda

Thank you, I'm actually working on it right now :)

Last edited by a moderator: May 7, 2017
15. Dec 23, 2014

### keithcuda

OK, so here is what I put together so far...

16. Dec 23, 2014

### Staff: Mentor

So far so good ....

17. Dec 23, 2014

### keithcuda

I guess this is where I am really stuck. Without a given angel for theta, how to I proceed with my problem?

18. Dec 23, 2014

### dean barry

g assumed at 9.81 m/s/s
Choose any incline angle (and therefore incline length) but dont exceed 14.764 ° incline though, as you wont have any force left to accelerate the mass up the incline.
The problem is, if the force is kept constant, at the top of the slope, the mass will still be moving (and accelerating).

So, you accelerate the mass so far up the slope then remove the (250 N) force and allow it to decelerate under gravity, to come to a rest at the top of the slope.
Ive ran a few examples through, and found that :
Regardless of the incline length, the length you apply the force over turns out to be the same.
The key is that you put the work in ( force * distance) which is equal to the final PE (m*g*h)
So :
250 * d = m * g * h
You may do the rest.

19. Dec 23, 2014

### Staff: Mentor

The problem includes a force 250N. Of the expressions you have derived, what can you equate to 250N?

20. Dec 23, 2014

### keithcuda

so, 250*d=100*9.8*5
d=4900/250
d=19.6