Calculating Net Force on Mass: 800N, 347N, 0.074m/s/s

AI Thread Summary
The discussion revolves around calculating the net force acting on a person being pulled by multiple ropes while considering their weight. Initially, there was confusion regarding the acceleration due to gravity, which was incorrectly stated as 0.074 m/s² but later corrected to 9.8 m/s². The participant calculated the resultant force from four different pulling forces, arriving at a net force of 347.01 N at a compass bearing of 85.91°. The importance of including weight in the calculations was questioned, highlighting the need for clarity in the problem statement. Ultimately, the net force and direction were determined based on vector summation of the forces applied.
mohlam12
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just curious...if someone was pulling me with a rope around my nech, he was exerting a force of 347N, and I weight 800N, what is the net force that is goign to be applied to my mass?? gravity is 0.074 m/s/s
because i think to get the net force i need the acceleration (Fn = ma) and here I have no acceleration...im just confused.
 
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Your question is not clear. Is the rope pulling straight up or at an angle?

Assuming that the rope is pulling straight up, and that by "here I have no acceleration" you mean you don't know the acceleration, then there is a force of 347N upward and a force of 800N downward, then the net force is simply 347-800= -453N (downward). You can then calculate the acceleration from Fn= ma.
 
What do you mean gravity is 0.074ms^{-2}...?Aren't you on Earth...?TO be strict,that #,namely ~9.8ms^{-2} should be called "acceleration due to gravity",or intensity of the Earth's gravitational field at the surface of the Earth...Terminology is not an issue in this case,but the origin of the figure you posted...:wink:

Daniel.
 
ok, here is the problem...im on earth, the gravity is 9.8 m/s/s
i got the highest score in physics class, my classmates got mad and put ropes around my neck and pulled me to the direction with different forces :
710.50 N 54deg CB
784.98 N 108deg CB
730.10 N 208deg CB
802.62 N 308eg CB

i weight 799.78 N

i calculated the 4 forces, and I got that the sum is 347.01N at 85.91deg CB

now the question is: what net dorce would they apply to your mass and in what direction?
i got the direction 85.94deg CB
now, what is the net force?

(sorry i put the gravity 0.074 for the first post )
 
I don't know what does that CB stands for,it's better if you did...The net force should be the resulting force,which is the vector sum of the 4 forces.If you said that the sum is 347.01N,then that is its modulus,the direction you found to be 85.91°,so you got the whole answer.I don't know if it's the correct one,or not,you have to do the calculations,not us.And just because it's simple trigonometry & arithmetics,you have no excuse for wrong result...

Daniel.
 
CB, compass bearing
yeah, but I didnt use my weight in anything, why my weight was given then?
 
Your exact phrasing of the question was:"what net dorce would they apply to your mass and in what direction?".Is gravity one of your classmates...?Or perhaps,the question was different...?:wink:

Daniel.
 
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