Last Saturday was extremely windy while we were out for a jog/run. The wind was blowing from the west to east, while we were running north. Clearly, I felt the effect of the wind. But from what I remember about vectors, a force applied in one direction has a zero component in the perpendicular driection. So since I was moving north, I should not have been affected by the wind. What gives?
Re: A question regarding vectors
so that means the force of wind wouldnt give you extra thrust to run faster or retard you to get slower.
same way if you were runing along the wind then it would throw you sideways
Re: A question regarding vectors
this is my understand, which might be wrong and that is you are not living on a 2D map.
Re: A question regarding vectors
---> ^
The wind is pushing right and you are running upwards.
There's my crude drawing. The wind is blowing to the right, and you're running up. The wind wouldn't (in an ideal case) slow you down or speed you up, but it would push you to the right. To keep going straight, you would have to apply force to the left to counter the wind.
Re: A question regarding vectors
Last Saturday was extremely windy while we were out for a jog/run. The wind was blowing from the west to east, while we were running north. Clearly, I felt the effect of the wind. But from what I remember about vectors, a force applied in one direction has a zero component in the perpendicular driection. So since I was moving north, I should not have been affected by the wind. What gives?
It was not the wind but your effort that was acting diagonally. Although you were jogging South to North, you were actually pushing yourself in North-West direction. A component of your push (east-west component) was negating the west-east wind and keeping you where you are. The second component of your force (South-North) was moving you in North direction. That is why you were feeling more force on your body than you should've if there was no West-East wind.
Re: A question regarding vectors
I think Ghost and TLK captured it. TLK'sexplanation is the clearest! Thanks guys.
Re: A question regarding vectors
Oh physics, I wanna romance thee.
I love that little concept southie
Re: A question regarding vectors
^ credit Ghost and TLK - they came up with the explanation
Re: A question regarding vectors
Is it possible to quantify the slowing effect. Let us go with the following input:
target running speed - 5 miles per hour
wind speed - 30 miles per hour
weight of person: xx lbs (use whatever you want - answer may vary with weight)
height of person:yy inches ((use whatever you want - answer may vary with wt and ht)
Atmospheric conditions - sea level - 14.7 psia
So what is the absolute magnitude and direction of the velocity of the runner if there was no wind.
(I dont know the answer - so this is not a quiz, just wanted to see what folks come up with).
Re: A question regarding vectors
I meant - what is the effort the runner has to put out and in which direction, such that with the wind, he goes at 5 mph South to North, with wind blowing west to east at 30 mph.
Re: A question regarding vectors
I tried analyzing this using the kinetic energy equation (K.E. = 1/2 mv^2) where m is the mass of the runner, and v is the velocity (for simplicity, I converted mph to m/s i.e S.I units)
5 mph is 2.24 m/s
30 mph is 13.41 m/s
Also I assumed the runner as a point object at his center of gravity (CG)
So in the baseline case (i.e no wind), the work done = K.E. = 1/2m(2.24*2.24) = 2.5m i.e. 2.5 times mass of the runner
Now considering the effect of the wind - the total work done will be the sum of work done on both axes (i.e North-South and East-West).
So work done = 1/2m(2.24*2.24) + 1/2m(13.41*13.41) = 2.5m + 90m (approx) = 92.5m
Now for simplicity assume mass of the person is 100 kg.
So in case 1, work done is 2.5 X 100 = 250 Joules
Case 2, work done is 9250 Joules
It's been a few years since I opened a Physics text book, so feel free to shoot holes in the above calculation :p
Re: A question regarding vectors
^ nice try LucyMay. Please go to the quantitative solution to vector problem thread. I have put my 2 paisa worth solution there. Your turn to shoot holes!
Just from a sniff test perspective, 9250/250 seems an excessively high ratio with wind vs w/o wind.
:p
Re: A question regarding vectors
I tried analyzing this using the kinetic energy equation (K.E. = 1/2 mv^2) where m is the mass of the runner, and v is the velocity (for simplicity, I converted mph to m/s i.e S.I units)
5 mph is 2.24 m/s 30 mph is 13.41 m/s
Also I assumed the runner as a point object at his center of gravity (CG)
So in the baseline case (i.e no wind), the work done = K.E. = 1/2m(2.24*2.24) = 2.5m i.e. 2.5 times mass of the runner
Now considering the effect of the wind - the total work done will be the sum of work done on both axes (i.e North-South and East-West). So work done = 1/2m(2.24*2.24) + 1/2m(13.41*13.41) = 2.5m + 90m (approx) = 92.5m Now for simplicity assume mass of the person is 100 kg.
So in case 1, work done is 2.5 X 100 = 250 Joules Case 2, work done is 9250 Joules
It's been a few years since I opened a Physics text book, so feel free to shoot holes in the above calculation :p
Work Done = Force X Dist. He is not covering any dist. in east-west direction work done is zero in that direction :D