Calculating mass given maximum tension of a rope

AI Thread Summary
To determine the maximum mass that can be lifted with a string capable of handling 85 N of tension while accelerating at 2.0 m/s² on the moon, the correct approach involves applying Newton's second law. The initial calculations led to a negative mass due to incorrect handling of forces and accelerations. By correctly setting up the equation ΣF = Fg + Ft = ma, and substituting the gravitational force (Fg = mg) and tension (Ft), the resulting mass was calculated to be approximately 24 kg. The discussion emphasized the importance of correctly applying signed values and understanding the direction of forces. Accurate calculations and a clear understanding of the physics principles are essential for solving such problems.
RohanTalkad
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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]
 
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RohanTalkad said:
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?
 
haruspex said:
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)
 
RohanTalkad said:
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 Ft and mg into that.
 
haruspex said:
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.

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?
 
RohanTalkad said:
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.
 
haruspex said:
You changed 1.6 to 1.5, but otherwise that looks right.

oooooh sorry :( but anyway thanks a million!
 
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