- #1
awelex
- 44
- 0
Hi,
here's a problem I did earlier today. I think I got it right, but I'd appreciate it if somebody could check. Also, is there a different way to solve this problem?
Two masses Ma and Mb are connected through a heavy rope with mass Mr. The top mass Ma is hanging from a from a massless rope that is that is attached to a helicopter that moves upwards with acceleration a. Find the tension in the top rope as well as the tensions in the bottom (heavy) rope.
I labeled the tension of the top rope T1, the tension of the heavy rope pulling downward on Ma T2.1, and the tension pulling upward on Mb T2.2.
For mass A: Fnet = Ma * a
--> T1 - T2.1 - Ma * g = Ma * a
--> T1 - T2.1 = Ma * a + Ma * g
For mass B: Fnet = Mb * a
--> T2.2 - Mb * g = Mb * a
--> T2.2 = Mb * a + Mb * g
This already gives us the tension from the heavy rope pulling upwards on the second mass.
For the entire system, without the top rope:
Fnet = (Ma + Mb + Mr) * a
--> T1 - (Ma + Mb + Mr) * g = (Ma + Mb + Mr) * a
--> T1 = (Ma + Mb + Mr) * (a + g)
I plugged this into the earlier equation with T1 and T2.1:
(Ma + Mb + Mr) * (a + g) - T2.1 = Ma * (a + g)
--> T2.1 = (Mb + Mr) * (a + g)
So we have:
T1 = (Ma + Mb + mr) * (a + g)
T2.1 = (Mb + Mr) * (a + g)
T2.2 = Mb * (a + g)
See above.
Is this correct?
here's a problem I did earlier today. I think I got it right, but I'd appreciate it if somebody could check. Also, is there a different way to solve this problem?
Homework Statement
Two masses Ma and Mb are connected through a heavy rope with mass Mr. The top mass Ma is hanging from a from a massless rope that is that is attached to a helicopter that moves upwards with acceleration a. Find the tension in the top rope as well as the tensions in the bottom (heavy) rope.
I labeled the tension of the top rope T1, the tension of the heavy rope pulling downward on Ma T2.1, and the tension pulling upward on Mb T2.2.
Homework Equations
For mass A: Fnet = Ma * a
--> T1 - T2.1 - Ma * g = Ma * a
--> T1 - T2.1 = Ma * a + Ma * g
For mass B: Fnet = Mb * a
--> T2.2 - Mb * g = Mb * a
--> T2.2 = Mb * a + Mb * g
This already gives us the tension from the heavy rope pulling upwards on the second mass.
For the entire system, without the top rope:
Fnet = (Ma + Mb + Mr) * a
--> T1 - (Ma + Mb + Mr) * g = (Ma + Mb + Mr) * a
--> T1 = (Ma + Mb + Mr) * (a + g)
I plugged this into the earlier equation with T1 and T2.1:
(Ma + Mb + Mr) * (a + g) - T2.1 = Ma * (a + g)
--> T2.1 = (Mb + Mr) * (a + g)
So we have:
T1 = (Ma + Mb + mr) * (a + g)
T2.1 = (Mb + Mr) * (a + g)
T2.2 = Mb * (a + g)
The Attempt at a Solution
See above.
Is this correct?