I don't understand f=ma So far so good

[Rosemann]

Alright so f=ma

I know the difference between mass and weight so that's alright. Now I'm heavy (96Kg); If I'm moving forwards at a rate of one metre per second (with the passing of every second I move forward exactly one metre) then is 'a' equal to one?

This confuses me in that I'm not accelerating. The rate of change of my velocity is zero - is it not?

My velocity is constant.

Or is that the same as saying my rate of acceleration is one?

f=ma so the force required to cause me to move forwards one metre per second per second is 96N Correct?

Work done is equal to force multiplied by distance and power (rate of working) in Watts is work multiplied by time. So If I move myself forwards one metre in the time frame of one second does that mean that the power required to do so was 96W?

Or is that I weigh 96Kg with Earth's gravitational constant of 9.81 in place therefore my mass is 96/9.81 = 9.89Kg. Surely that can't be right can it?

 

[Spinnor]

If you are moving with constant velocity your acceleration is 0.

 

[Fredrik]

In addition to what spinnor said, you also need to keep in mind that F and a are vectors. They both have a direction, and the equality sign ensures that it's the same direction.

 

[Ocasta]

Alright so f=ma
This confuses me in that I'm not accelerating. The rate of change of my velocity is zero - is it not?

Your acceleration is your change in velocity over time. So if you started at v₀ = 5 m/s and accelerated to v= 6 m/s you would have acceleration. If you constantly move 5 m/s, you have acceleration = 0.

So if we plugged this into the equation,
$Force = mass (0)$

If you have zero acceleration, you are exerting zero force. But bear in mind that if you are running around on Earth, you need to exert force in order to move. If, however, you're between galaxies and moving at a constant velocity, you are exerting no force.

 

[Rosemann]

So if we plugged this into the equation,
$Force = mass (0)$

If you have zero acceleration, you are exerting zero force. But bear in mind that if you are running around on Earth, you need to exert force in order to move. If, however, you're between galaxies and moving at a constant velocity, you are exerting no force.

Thank you all for the replies.

I can see the galaxy idea because in effect there is no resistance to my movement which will slow me down and so inertia (momentum) keeps me going along at a constant velocity i.e. no extra force is required to keep me moving.

So then, I'm either close to understanding what's going on here or else finally completely burning out the three remaining neurons l have left!

If I have a toy car on the coffee table and I push it forward with my finger at a constant velocity then the acceleration seen by the toy is zero. That being the case f=m.(0) so the force being applied to the car to move it forward is zero i.e. there isn't any. So if no force is being applied to the toy why does it move?

If I stop pushing (applying a directional (vector) force to the car) it will stop moving forwards so clearly there is a force being applied but the equation says the magnitude of said force is zero, how can that possibly be?

 

[Fredrik]

F is the sum of all the forces acting on the object whose acceleration is a. The force that your hand exerts on the car is obviously non-zero, and in the forward direction. But that's not the only force acting on the car. Can you think of a force that pulls it in the opposite direction?

 

[Rosemann]

This sounds as though it has something to do with, 'Every action has an equal and opposite reaction'. So therefore... I need to think about this...

 

[Fredrik]

That "every action..." stuff is telling you that when object 1 is exerting a force on object 2, then object 2 is also exerting a force on object 1, which is equal in magnitude but has the opposite direction. The answer to your F=ma question doesn't have anything to do with that. F=ma is about the acceleration of a single object, and the forces that other objects exerts on it. So the force that the car exerts on your hand is irrelevant.

The answer you're looking for is "friction". That's the force (exerted on the toy car by the table) that exactly cancels the force that your hand exerts on the toy car.

 

[Rosemann]

I see, that makes sense. So the f=ma in this instance simply won't help me to calculate the force being applied by my finger to the car - is that correct?

If that is the case then is there a simply way to calculate said force?

 

[Fredrik]

That's correct. There's no way to calculate that force in this scenario, unless you have already done other experiments to determine the friction between the toy car and the table at the speed we're talking about. But you could measure the force, by pushing the toy car with a spring that gets a compressed a little, instead of pushing it with your finger.