How Is Tension Calculated in a Lifting Cable?

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In summary, the load was lifted 21.0m in 1.50 seconds, and the tension in the cable was 3040 N. The net work done on the load was 8330 J, and the work done by gravity on the load was -3040 N. The final speed of the load was v_f = (3040 N)(0.150 g) + 8330 J = 11960 J.
  • #1
rperez1
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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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 

Related to How Is Tension Calculated in a Lifting Cable?

Q: What is the weight of the load being lifted?

The weight of the load being lifted is 270kg.

Q: How is the load being lifted?

The load is being lifted using a lifting mechanism such as a crane or a forklift.

Q: What is the maximum weight that can be lifted by the lifting mechanism?

The maximum weight that can be lifted by the lifting mechanism depends on its capacity, which varies depending on the type and model of the mechanism. However, in this scenario, the maximum weight that can be lifted is 270kg.

Q: What is the force required to lift the load?

The force required to lift the load is equal to its weight, which is 270kg multiplied by the acceleration due to gravity (9.8 m/s^2). This results in a force of 2646 N.

Q: What is the potential energy gained by lifting the load to a certain height?

The potential energy gained by lifting the load to a certain height can be calculated using the formula PE = mgh, where m is the mass of the load (270kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height to which the load is lifted. For example, if the load is lifted to a height of 10 meters, the potential energy gained would be 26460 J.

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