# Calculating mass given maximum tension of a rope

1. Nov 3, 2015

If a string can handle a max tension of 85 N what is the largest mass that could be lifted with this string if it is lifted with an acceleration of 2.0 m/s^2 up on the moon?

Let up be positive in value.

Ft = - 85 N, a = 2.0 m/s^2, Fg = -1.6 m/s^2

2. Relevant equations

F = ma -> Fg = mg -> Ft = ma + mg -> Ft = m ( a + g)

3. The attempt at a solution

When I do the problem, I get a negative mass ...

Ft = m (a + g) -> -85 N = m ( 2 m/s^2 - 1.6 m/s^2 ) -> m = - 212.5 kg

Do you see the mistake?

Last edited: Nov 3, 2015
2. Nov 3, 2015

### haruspex

By what logic do you add these accelerations? Is it accelerating at rate a faster than gravity?
Draw a free body diagram. What are the forces acting on the body? What is the resulting acceleration?

3. Nov 3, 2015

Oh sorry. In the downward direction there is Fg and Ft.

By what logic do you add these accelerations? Is it accelerating at rate a faster than gravity?

Hm, I guess I added them because they were in two different directions (the lifting of the string was upward and the gravity as downward, but I realize that this is probably a mistake right?)

Oh, and by the way the orignal equation I did was Ft = mg + ma so that lead to Ft = m ( a + g)

4. Nov 3, 2015

### haruspex

I understand that seems intuitively right, but it fails when you plug in the signed values.
Remember, the basic equation is $\Sigma F_i = ma$. Plug Ft and mg into that.

5. Nov 3, 2015

I will try to do what you said

∑F = Fg + Ft = ma

And Fg = mg

so (mg) + Ft = ma

m (-1.6 m/s^2) - 85 N = m (2 m/s^2)

-1.6 m - 85 = 2 m

m = 23.611111111 kg -> 24 kg

how does that look?

6. Nov 3, 2015

### haruspex

You changed 1.6 to 1.5, but otherwise that looks right.

7. Nov 3, 2015