- #1
Mr Davis 97
- 1,462
- 44
I have a question about sign when it comes to solving vector problems. Here is one for example:
A 5.0-kilogram mass is positioned on a 30 degree incline. It is kept stationary by a string pulling parallel to the incline. Determine the tension of the string.
This is a very easy problem. I started by defining down the incline as positive and up the incline as negative. Thus,
##F_{net} = F_{g\parallel } - T##
##0 = F_{g\parallel } - T##
## T = F_{g\parallel } ##
##T = mg\sin\theta##
##T = 25 N##
I am getting 25 Newtons, not -25 Newtons. Why is this? The force of the tension, which is in the opposite direction as the parallel component of gravity (+25 N), should be negative, since it is cancelling the positive force. Where am I going wrong in the calculation?
A 5.0-kilogram mass is positioned on a 30 degree incline. It is kept stationary by a string pulling parallel to the incline. Determine the tension of the string.
This is a very easy problem. I started by defining down the incline as positive and up the incline as negative. Thus,
##F_{net} = F_{g\parallel } - T##
##0 = F_{g\parallel } - T##
## T = F_{g\parallel } ##
##T = mg\sin\theta##
##T = 25 N##
I am getting 25 Newtons, not -25 Newtons. Why is this? The force of the tension, which is in the opposite direction as the parallel component of gravity (+25 N), should be negative, since it is cancelling the positive force. Where am I going wrong in the calculation?