How Do You Calculate the Net Force on a Ball in Flight?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 9K views
Panzer22
Messages
2
Reaction score
0
1. A ball is flying through the air, and at a given point in the flight, the ball is submitted to a gravitational force: F(g) = 1,5 N [downwards] and air resistance: F(air) = 0,50 N [32° above the horizontal]. Calculate the net force on the ball.


2. F (nette) = m x a



3. I thought I might have to find the vertical and horizontal components of the air resistance, so I found:
sin32 x 0,5N = 0,26N = x
cos32 x 0,5N = 0,42N = y

I also tried calculating the mass of the ball, in case it might prove useful:
Fg = m x g
1,5N = 9,8m
0,153 kg = m


But after calculating these values, I'm at a loss of what to do next. How do you find out the net force without having the other force on the horizontal axis? How can I use this data to calculate the net force? The answer key in my book says it should be 1,3N [71° under the horizontal], but I have no idea how this answer was achieved. I would be grateful for any help in solving this problem.
 
Physics news on Phys.org
Panzer22 said:
I thought I might have to find the vertical and horizontal components of the air resistance, so I found:
sin32 x 0,5N = 0,26N = x
cos32 x 0,5N = 0,42N = y
You have the x and y components interchanged.

How do you find out the net force without having the other force on the horizontal axis?
Why should it have a non-zero horizontal component?

How can I use this data to calculate the net force?
Think of it this way: "Downwards" is 270 degrees away from the positive x-axis. Can you find the components of the gravitational force now?
 
neutrino said:
Why should it have a non-zero horizontal component?

If the ball is in movement, shouldn't the horixontal component be non-zero?

And what does it being 270 degrees away from the positive x-axis have to do with it?
 
To find the net force just add the two force vectors. You know what their components are along each direction.

And what does it being 270 degrees away from the positive x-axis have to do with it?
I was just trying to show you that the horizontal component of the gravitational force is zero.