Truck and Man Forces in Boat Tension Calculations

[Teenytiny1991]

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

 

[Teenytiny1991]

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

 

[PhanthomJay]

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.

 

[JazzFusion]

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.

 

[Teenytiny1991]

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

 

[Teenytiny1991]

P= person by the way =)

 

[PhanthomJay]

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.

 

[Teenytiny1991]

does it mean there is no net force. so does that mean it would be zero?

 

[PhanthomJay]

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?

 

[Teenytiny1991]

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

 

[PhanthomJay]

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.