What is the tension in the cable

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Homework Help Overview

The discussion revolves around calculating tension in a cable during a rescue operation and analyzing forces acting on a moving object. The first part involves determining the tension when a man is lifted with an initial acceleration and then at constant velocity. The second part focuses on finding the magnitude and direction of a third force required for an object to maintain constant velocity.

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Approaches and Questions Raised

  • Participants discuss the calculation of tension in the cable based on weight and acceleration. Questions arise regarding the direction of forces and the application of trigonometric functions to find angles in force diagrams.

Discussion Status

Some participants have provided calculations and attempted to clarify their reasoning. There is ongoing exploration of the correct method for determining angles in force analysis, with various interpretations being discussed. Guidance has been offered regarding the use of trigonometric functions, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also a noted confusion regarding the application of trigonometric functions in the context of force direction.

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4) A rescue helicopter is lifting a man (weight- 822 N) from a capsized boat by the means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of 1.10m/s^2? (b) What is the tension during the remainder of the rescue when he is pulled upward at constant velocity?

a) The mass of the man is 83.9kg

T = (83.9Kg)(9.80m/s^2+1.10m/s^2)

T = 914.5N

b) It is 822N because the velocity is constant.


5) Three forces act on a moving object. One force has a magnitude of 80.0N and is directed due norht. Another has a magnitude of 60.0N and is directed due west. What must be the magnitude and direction of the third force, such as the object continues to move with constant velocity.

I know how to find the magnitude, but have trouble finding the direction. Someone suggested tan(angle)= x/y. But when I do it out out i get a small number that couldn't possibly be the angle.

Thanks
 
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You did problem 4 OK.
Originally posted by physicszman

I know how to find the magnitude, but have trouble finding the direction. Someone suggested tan(angle)= x/y. But when I do it out out i get a small number that couldn't possibly be the angle.
Show what you did. (You must have made an error somewhere!)
 
Hi, thanks for your help.

I think I realized what I did wrong. You have to turn the formula around to solve for just the angle so its

angle = (80N/60N)/tan = 57.2

Do I answer the direction is 57.2 degrees North West?
 
angle = (80N/60N)/tan = 57.2

Thats how I got the previous answer. How do I calculate it the right way?

And I guess it would be from the X axis since the y-component of the force is greater.
 
Last edited:
Last question..

So what do I do with the 4/3 take the "arctan" of it?

Thanks
 
53.1 degrees from west axis. I hope this sounds right. Thanks again for all your help!
 

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