A firefighter who weighs 700N slides down a vertical pole w/ an acceleration of 3.00m/s^2 directed downward. What is the magnitude of the frictional force of the pole on the firefighter?
To whomever is out there tonight. Thank you, for any help while I try to study for my physics exam.
Finding magnitude of frictional force.?
2015-10-20 4:14 am
回答 (2)
2015-10-20 4:27 am
✔ 最佳答案
Remember sum of forces in y direction. We are told that they sum with an acceleration of 3m/s2 downward. I am declaring downward direction as positive values and upward direction as negative values.Friction is a force slowing the firefighter down so it is acting upwards.
I would say that you are probably using 9.8m/s2 but I am going to use 10m/s2.
700N = 700kg*m/s2
700kg*m/s2 / 10m/s2 = 70kg
70kg*3m/s2 = 210kg*m/s2
210kg*m/s2 = 700N + (- friction)
210N + friction = 700N
I conclude the friction has a magnitude of 490N.
2015-10-20 4:39 am
Accelerating force Fa = m*a = Fw/g*a = 700*3/9.806 Newton
Friction force Ff = weight force Fw minus accelerating force Fa
Ff = 700-700*3/9.806 = 700(1-3/9.806) = 486 N approx.
Friction force Ff = weight force Fw minus accelerating force Fa
Ff = 700-700*3/9.806 = 700(1-3/9.806) = 486 N approx.
收錄日期: 2021-04-21 14:40:45
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