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donotremember
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To keep Robin from being captured, Batman tosses him out of a third-story window, knowing that a 17.0m rope hangs slack between hooks of equal height on adjacent buildings 13.0m apart. Robin grabs the rope and hangs on at a point 5.0m from one end. Assuming that Robin's mass is 45.0kg and the rope withstands the initial impulse, what is the tension in each part of the rope when equilibrium is established?
I believe that this question can be simplified to the following:
(Think of this as a triangle)
Find T2 and T1:
.......13m
\---------------------------------------------/
...\........../
..5m..\......./
....\.../...12m
.....\-/
......|
......|
......|
......V
......45kg or (45kg)(9.8m/s^2) = 441NWhere the 5m rope is T2 and the 12m rope is T1
From the law of cosines we have the following angles
67.4 (or 112.6).........22.6
\---------------------------------------------/
...\........./
...\....../
...\.../
...112.6...\-/...22.6
---------------------------------------------------------
Step 1 find a relation between T2 and T1:
The system is in equilibrium so the x components cancel
T2cos112.6 + T1cos22.6 = 0
T2(-0.384) = -T1(0.923)
T2 = 2.4T1
Step 2 use the relation found in step one to solve for T1
The system is in equilibrium so the y components added equal the weight:
T2sin112.6 + T1sin22.6 = 441N
2.4T1sin112.6 + T1sin22.6 = 441N
2.215T1 + 0.384T1 = 441N
T1 = 169.6N
and given step 1 T2 = 407N
The answers given in my book are
T1 = 1.5 X 10^2 N
T2 = 3.6 X 10^2 N
Note that they satisfy the relation T2 = 2.4T1
Also note that gravity is calculated as being 9.8m/s^2 in my book (which I have used)
I believe that this question can be simplified to the following:
(Think of this as a triangle)
Find T2 and T1:
.......13m
\---------------------------------------------/
...\........../
..5m..\......./
....\.../...12m
.....\-/
......|
......|
......|
......V
......45kg or (45kg)(9.8m/s^2) = 441NWhere the 5m rope is T2 and the 12m rope is T1
From the law of cosines we have the following angles
67.4 (or 112.6).........22.6
\---------------------------------------------/
...\........./
...\....../
...\.../
...112.6...\-/...22.6
---------------------------------------------------------
Step 1 find a relation between T2 and T1:
The system is in equilibrium so the x components cancel
T2cos112.6 + T1cos22.6 = 0
T2(-0.384) = -T1(0.923)
T2 = 2.4T1
Step 2 use the relation found in step one to solve for T1
The system is in equilibrium so the y components added equal the weight:
T2sin112.6 + T1sin22.6 = 441N
2.4T1sin112.6 + T1sin22.6 = 441N
2.215T1 + 0.384T1 = 441N
T1 = 169.6N
and given step 1 T2 = 407N
The answers given in my book are
T1 = 1.5 X 10^2 N
T2 = 3.6 X 10^2 N
Note that they satisfy the relation T2 = 2.4T1
Also note that gravity is calculated as being 9.8m/s^2 in my book (which I have used)
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