# Homework Help: Simple Newton's 2nd Law problem Why isn't this answer right, though?

1. Dec 12, 2012

### Lo.Lee.Ta.

1. "A 40.0-kg crate is pulled along a horizontal floor by the ideal arrangement
shown in figure below. The force F is 280 N. The coefficient of kinetic friction between the crate and the floor is 0.170. What is the acceleration of the crate?"

______box---^=Pulley__

Okay, this is a weird drawing of the situation. Just a box on a flat surface.
It's attached by a string to a pulley and the other end of the string is anchored to a
cone (^).

So I thought the solution is just this:

εF(y) = ma(y)

-w + N = ma(y)
-392 + N = (40)(0)
N= 392N

εF(x) = ma(x)
-f + F = ma(x)
[mu(k)](N) + 280 = (40)(a)
(.17)(-392) + 280 = (40)(a)
a= 5.33m/s^2

BUT, this is not the right answer on my test!
How do you get that answer???

Thank you so much! :D

2. Dec 12, 2012

### Staff: Mentor

Can you post the diagram? It's not clear what the arrangement is.

3. Dec 12, 2012

### Lo.Lee.Ta.

Okay, this is it.

4. Dec 12, 2012

### Staff: Mentor

Okay. If the pulley is massless, how is the tension in the cord related to the force F?

(a) Tension = F
(b) Tension = F/2

5. Dec 12, 2012

### Staff: Mentor

Good. So now figure out the tension in the rope. That will tell you the force on the crate.

6. Dec 12, 2012

### Lo.Lee.Ta.

Um, okay...

So I looked up the formula for tension in a rope.

They say it's: F(tens.) = ma - (mu)(-mg)

To find the a, I did: F = ma
280 = (40)(a)
a = 7m/s^2

So, that means it's: F(tens.) = (40)(7) - (.17)(-392)
F(tension) = 346.64N

So, then I thought I would use the tension force instead of the F...

-f + T = ma(x)

-392 + 346.64 = (40)(a)
a = -1.134m/s^2 BUT THIS IS NOT RIGHT...

It's negative. I don't really know what's wrong. I'm thinking the F(tension) is wrong... =_=

7. Dec 12, 2012

### Staff: Mentor

To find the tension in the rope analyze the forces acting on that massless pulley.

8. Dec 12, 2012

### Lo.Lee.Ta.

It seems to me that the tension would be equal to the force, F.

Pulley:
ƩF(x) = ma(x)
-T + 280 = (0)(a)
T = 280

So I still don't see why the Tension isn't the same thing as the Force F...

Box:
ƩF(x) = ma(x)
-f + T = ma(x)
(.17)(-392) + 280 = 40(a)
a = 5.334m/s^2.... The same wrong answer... Is there some force on the Pulley I'm leaving out?!
It is massless, so there would be no N force or weight.

9. Dec 12, 2012

### ap123

If you look at the pulley, then you have two tension forces acting to the left.
So, 2T = F
→ T = 140N

10. Dec 12, 2012

### Lo.Lee.Ta.

WOOO! Hallelujah!

Thanks, ap123!!! I've been working on this problem FOR-EVER!!!
But what the heck?!
2 tensions?
Is this because of the tension where the string attaches to the box AND where the string from the pulley attaches to the cone?
-Both are pointing in the left direction.

11. Dec 12, 2012

### ap123

Yes.

Last edited: Dec 12, 2012