How Is Tension Calculated in a Lifting Cable?

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The discussion focuses on calculating tension in a lifting cable for a 270-kg load accelerated at 0.150 g over a distance of 21.0 m. The tension in the cable was determined to be 3040 N, and the net work done on the load was calculated as 8330 J. Participants discussed methods for solving the remaining parts of the problem, specifically the work done by gravity and the final speed of the load. For work done by gravity, the equation W = -mg(h) was suggested, emphasizing the negative sign due to gravity opposing the lift. To find the final speed, basic kinematics and integration of acceleration were recommended to determine the time taken to lift the load.
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Homework Statement


A 270-kg load is lifted 21.0 m vertically with an acceleration a=0.150 g by a single cable.

A)Determine the tension in the cable.
B)Determine the net work done on the load.
C)Determine the work done by gravity on the load.
D)Determine the final speed of the load assuming it started from rest.

Homework Equations

The Attempt at a Solution


i was able to get both A and B, for A my answer was...
FT =
3040 N

and for B the answer i got was
Wnet =
8330 J

for C and D I don't fully understand how to solve, any ideas? please help
 
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For part D, the speed is based on basic mechanics and your acceleration.
## v_f = \int_{t_0}^{t_f} a dt ##
Assuming t_0 = 0 and v_0 = 0, you will get an answer for v_f based on total time.
To find total time, you integrate once more, and solve for how long it takes to go 21m.

For part C, I am not too sure, but I would treat gravity like any other force and plug it into your equation for work.
One I found on the internet is W
0067.png?rev=2.5.png
= -mg(
0394.png?rev=2.5.png
h). The negative sign picked up since gravity is working against the distance of travel in this case.
 
RUber said:
For part D, the speed is based on basic mechanics and your acceleration.
## v_f = \int_{t_0}^{t_f} a dt ##
Assuming t_0 = 0 and v_0 = 0, you will get an answer for v_f based on total time.
To find total time, you integrate once more, and solve for how long it takes to go 21m.

For part C, I am not too sure, but I would treat gravity like any other force and plug it into your equation for work.
One I found on the internet is W
0067.png?rev=2.5.png
= -mg(
0394.png?rev=2.5.png
h). The negative sign picked up since gravity is working against the distance of travel in this case.
i am so sorry but i do not understand? could you elaborate?
 
rperez1 said:

Homework Statement


A 270-kg load is lifted 21.0 m vertically with an acceleration a=0.150 g by a single cable.

A)Determine the tension in the cable.
B)Determine the net work done on the load.
C)Determine the work done by gravity on the load.
D)Determine the final speed of the load assuming it started from rest.

Homework Equations



The Attempt at a Solution


i was able to get both A and B, for A my answer was...
FT =
3040 N

and for B the answer i got was
Wnet =
8330 J

for C and D I don't fully understand how to solve, any ideas? please help
How did you get the answers for A and for B ?

C is similar.

For D, use simple kinematics, or use the work-energy theorem.
 
rperez1 said:
i am so sorry but i do not understand? could you elaborate?
I only learned calc-based physics, so I am not sure how to be more clear.
Velocity is the integral of acceleration.
For a constant acceleration, you have ##v(t) = at + v_0##
In this case, v_0 is 0 because it says the object starts at rest.

Position is the integral of velocity.
For a linear velocity, like v = at, position ##p(t) = \frac a2 t^2 +p_0##.
In this case, you want to find the time it takes for p(t) = 21, with p_0 = 0.
This will tell you how long it takes to raise the object to 21m.

Once you have the time, you plug that into your linear equation for velocity to solve the problem.
 
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