Uniform circular motion problem

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In a uniform circular motion problem, a police officer driving at 70.9 km/h through a turn with a radius of 356 m must consider both horizontal and vertical forces to determine the net force acting on her. The centripetal force, calculated as 62.2 N, is directed toward the center of the circle, while the officer's weight also contributes to the net force. The discussion emphasizes the importance of vector addition in calculating the net force, which includes both the centripetal force and gravitational force. Participants clarify that the angle relative to vertical must be determined by considering the resultant vector of these forces. The conversation highlights the need for careful interpretation of force directions and the correct application of trigonometric methods to find angles.
  • #31
Nathanael said:
FT is the net force right? The magnitude of FT is not 559.98.

Also there are 2 angles against the vertical. The problem may want you to find the smaller of the two angles.
i guess i found it.
20150627_004414_zpssedy1fr0.jpg
 
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  • #32
Nathanael said:
FT is the net force right? The magnitude of FT is not 559.98.

Also there are 2 angles against the vertical. The problem may want you to find the smaller of the two angles.
thanks again Nathanael!
 
  • #33
No problem. I was going to tell you about the arctan method after you found it your way. The arctan method was the first that came to my mind.
(But the arccos(a.b/(|a||b|)) method is also a very good one to know! It's far more general.)
 
  • #34
i really need to work on recognizing what is the logical next step to take with what information i have. i spend way too much time on a single problem
 
  • #35
Nathanael said:
No problem. I was going to tell you about the arctan method after you found it your way. The arctan method was the first that came to my mind.
(But the arccos(a.b/(|a||b|)) method is also a very good one to know! It's far more general.)
did you find the angle with the arccos method? what values did you use for a and b? i could only get 180degs
 
  • #36
J-dizzal said:
i really need to work on recognizing what is the logical next step to take with what information i have. i spend way too much time on a single problem
It comes with practice. I hope at least that you feel you are getting better and better (even if slowly).

J-dizzal said:
did you find the angle with the arccos method? what values did you use for a and b? i could only get 180degs
I didn't use that method, but it would work. You had the dot product right (62\hat i-mg\hat j)\cdot(0\hat i+\hat j)=-mg but then you are supposed to divide by \sqrt{(mg)^2+62^2}*1 but for some reason you were dividing by mg. That's why in post #30 I said the magnitude of the net force is not mg.

I think it was just a careless mistake, just got to take things slow.
 
  • #37
Nathanael said:
It comes with practice. I hope at least that you feel you are getting better and better (even if slowly).I didn't use that method, but it would work. You had the dot product right (62\hat i-mg\hat j)\cdot(0\hat i+\hat j)=-mg but then you are supposed to divide by \sqrt{(mg)^2+62^2}*1 but for some reason you were dividing by mg. That's why in post #30 I said the magnitude of the net force is not mg.

I think it was just a careless mistake, just got to take things slow.
yep i sure did miss that, thanks for point it out to me though.
 
  • #38
ill be on tomorrow with some statics problems, most likely. hope youre on to help. goodnight. thanks again.
 
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