Truck and Man Forces in Boat Tension Calculations

  • Thread starter Teenytiny1991
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    Tension
In summary: You don't need to show your work, but you should post your calculations so someone else can check them.
  • #1
Teenytiny1991
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Homework Statement


A Pickup truck pulla a boat down a creek witha constant force of 860 N. A man helps steer the boatby pulling a rope from the opposite side of the creek. The man and the truck exert just enough force to keep the boat moving at a constant speed in the center of the creek. The angle from the bow to the truck is 22 degrees and the angle from the bow to the person is 75 degrees A) what i the tension in the man's rope B) what is the magnitude of the resistive force of the watr against the boat C) what is the resultant of all forces acting on the boat

I worked out the problems and got answers i was wondering if you can tell me if they are right or wrong?

Homework Equations





The Attempt at a Solution


A) t= 333.53 N
B) Fr=883.7N
C) 883.7 N
 
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  • #2
i have the work to go with them it w9ould just save me a lot of time if someone would just say yes or no. but if no one is willing to do that can you just wite on the post and tell me to post them so you can help me? please
 
  • #3
A sketch would be helpful. At first glance, there is not enough information to solve the problem. It appears that there are 2 unknowns, T and Fr, but only one equation to work with. That's why a sketch would help. I can tell you that part c is wrong, since if the boat moves at constant speed, what does Newton's law tell you about the net or resultant force? Please show your work.
 
  • #4
Um, isn't this one of our questions on the take-home quiz? I don't think it's right to ask for help on that.
 
  • #5
I don't know how to draw on here but here are my calculations

EFy=0=860sin22-Psin75
0=322.16-.966P
-322.16=-.966P
T=333.53N

EFx=0=860cos22+333.53cos75
797.38+86.32
Fr=883.7

Then i did pythagorean theorom to add the two vectors. Sorr i misprinted my answer for C
883.7^2+333.53^2 (in a square root)=944.55 N
 
  • #6
P= person by the way =)
 
  • #7
Sorry, I misinterpreted your problem. Your calc for the tension in the person's rope is correct. Your calc for the force of the water resistance is correct (although you set the sum of forces in x dir =0, actually, that's correct, but you didn't have the Fr term in there). But your answer to the resultant is wrong. You didn't include the tension from the pickup. But you don't have to go any further. If the boat is moving at constant speed, you know that the resultant force must be_____?___ . And BTW, it is perfectly acceptable to seek homework or exam help on this forum, as long as you show your work and relevant equations. Answers are not given out here: homework helpers help, and you provide the answers.
 
  • #8
does it mean there is no net force. so does that mean it would be zero?
 
  • #9
Well you already noted (I think you did, but maybe you didn't), that the sum of x forces was 0 and the sum of y forces was 0. So what does that imply about the net or resultant force? And why is the sum of the x forces and y forces equal to 0?
 
  • #10
the y forces equal zero because nothing is acting upon the boat in either the y or -y way. and to be honest the only reason i said EFx is equal to zero is because I was setting the equation equal to zero. as habit. I'm not sure why the X forces would equal zero.
And the accellaration of the boat is directly proportional to the net force acting on it. and the net force on an object = ma. but since the speed is constant there is 0 accellaration
 
  • #11
Teenytiny1991 said:
the y forces equal zero because nothing is acting upon the boat in either the y or -y way. and to be honest the only reason i said EFx is equal to zero is because I was setting the equation equal to zero. as habit. I'm not sure why the X forces would equal zero.
And the accellaration of the boat is directly proportional to the net force acting on it. and the net force on an object = ma. but since the speed is constant there is 0 accellaration
Yes, since the speed is constant, a=0 in the x direction; that's Newton 1. I like your honesty.
 

1. How do I know if my answer is correct when checking for tension?

The best way to check your answer for tension is to use a tension calculator or formula specific to the type of problem you are solving. You can also double-check your calculations and make sure you are using the correct units.

2. What should I do if my answer for tension doesn't match the expected value?

If your answer for tension does not match the expected value, first check your calculations for any errors. If you are confident in your calculations, it is possible that your answer is within an acceptable margin of error. In this case, you can state your answer with the appropriate precision and note the margin of error.

3. Is it important to check for tension when solving a problem?

Yes, it is important to check for tension when solving a problem. Tension is a crucial factor in many scientific and engineering calculations, and an incorrect answer can lead to errors and inaccuracies in your overall solution.

4. Can I use the same method to check for tension in different types of problems?

The method for checking tension may vary depending on the type of problem you are solving. It is best to use a tension calculator or formula specific to the problem at hand, as different materials and systems may have different tension calculations.

5. Are there any common mistakes to watch out for when checking for tension?

Some common mistakes to watch out for when checking for tension include incorrect unit conversions, using the wrong formula or calculator, and not accounting for all the necessary variables in the problem. It is important to double-check your work and make sure all the necessary factors are included in your calculations.

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