Calculating mass given maximum tension of a rope

[RohanTalkad]

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

Homework Equations

F = ma -> Fg = mg -> Ft = ma + mg -> Ft = m ( a + g) [/B]

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? [/B]

 

[haruspex]

a + g

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?

 

[RohanTalkad]

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?

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)

 

[haruspex]

Ft = mg + ma

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 F_t and mg into that.

 

[RohanTalkad]

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 F_t and mg into that.

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?

 

[haruspex]

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)

-1.5 m - 85 = 2 m

m = 24.2857 kg -> 24 kg

how does that look?

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

 

[RohanTalkad]

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

oooooh sorry :( but anyway thanks a million!