0
$\begingroup$

This is a projectile motion question and the question basically states that an object is launched at $45$ degrees with an initial velocity of $28m/s$ east.

So, how many forces are acting on a projectile when it is passing through the highest point in the trajectory if air resistance is ignored?

According to my textbook answer to a specific question:

If air resistance is ignored, the only force acting is gravity, Fg, and so the net force will be: $Fg = mg = 80 × 9.8 = 784 N$ down

Okay $F_g$ is acting downwards on the object, fine I get this. But isn't the object also moving towards east at constant speed of $28\cos \left(45\right)m/s$ , so shouldn't this mean that the object also has a horizontal force applied to it?

$\endgroup$
4
  • 2
    $\begingroup$ I'll also note that there's nothing unique about the peak of the trajectory - gravity is the only force acting on the object for duration of the flight. $\endgroup$ Sep 11 '20 at 13:10
  • $\begingroup$ It's an absolutely KEY concept in Newtonian mechanics that if an object is traveling at constant velocity, NO net force is acting on it. This means that when you are driving down the road in a straight line in your car, at a constant speed of 60 miles/hr, the force that your engine is applying to your tires to keep your car moving is EXACTLY balanced by wind drag and other friction forces, leading to ZERO net force on your vehicle. This concept MUST be internalized and believed if you are to make progress in learning how to do Newtonian mechanics. $\endgroup$ Sep 11 '20 at 21:09
  • $\begingroup$ @NuclearWang Thanks for the note. I have one more question, is the velocity of the projectile tangent to the parabolic path of the projectile? $\endgroup$
    – Sarah V.P
    Sep 12 '20 at 8:11
  • $\begingroup$ @DavidWhite That's a partial answer, not a comment. Flesh it out and post it. $\endgroup$
    – Bill N
    Sep 17 '20 at 1:05
2
$\begingroup$

Forces describe the change of velocity (or momentum, technically speaking) of an object. Not the current velocity, just the change. If the velocity towards east doesn't change (= is constant), then there can't be any net force applied to the object in that direction. The direction in which velocity changes is the vertical one, so the net force applied to the projectile must be in that direction as well, and gravity is fully sufficient to describe this vertical force.

$\endgroup$
1
$\begingroup$

The object is moving towards the east with a constant velocity at all points of its trajectory and this is not due to a force acting on it but due to no force acting on it in that direction (ignoring air resistance).

Let me clarify.

enter image description here

You projected the object which was at rest earlier with some velocity at $45°$ . This means you applied a force to throw the object but the moment the object left the contact with your hand, there is no force acting on it except the downward gravity and because of this force of gravity , the vertical component of the velocity given changes but since there is nothing to apply a force on the object in the horizontal direction , it's velocity in that direction doesn't change.

So we don't need to apply force to move a body with constant velocity.

There was a similar confusion with Aristotle who said that we need to apply a force to a body to make it move constantly . But Galileo then said that we don't need to apply a force to a body to move it constantly but a force is needed if we want to change the state of the body i.e. to make it move from rest or change the velocity with which it was moving earlier.

Hope it helps 🙂.

$\endgroup$
1
$\begingroup$

The only force acting on the particle thoughout its trajectory is gravity.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.