This great and super helpful. I appreciate it! I think after thinking about the rolling resistance and aerodynamic drag my answer is pretty well answered.
I apologies, I used a YouTube video to figure out what forces I should include and friction was added into their equations. I wanted to add it to be safe but I made a mistake adding them without explanation. Friction between the tires and the road is what I meant. Once again, I assumed...
I realized that my free body diagram comment was misstated. The F21 and F12 would cancel out within the system Fnet Force equation because they are equal and opposite (however, acting on different objects). You would still include it when describing the forces on each object in their free body...
Okay so I shouldn’t have included friction in either case or only car 2 has friction included?
I believe that’s called the traction force (ut*N at maximum) and yes, I made a mistake. Therefore,
Fnetsys = Fgc - F21 - Ff2 + F12 or Fnetsys = Fgc - F21 + F12
I made this up and solved this problem:
I’m going to change the carts to cars so now we have car 1 and car 2.
Car 1’s mass is 1,300 Kg and car 2’s mass is 1,700 kg. There is a coefficient of friction of 0.3.
Car 1 is applying a force on the ground and the ground applies a force on car 1...
Newton's first law: An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force.I may have found my confusion. If the action-reaction pair is an internal force within the system that is the cars. Then, the external force...
I wanted to take some time to think a bit and decided to give more details.
It’s easiest to see the force applied by horse’s foot vs. it’s back will be different because of the control a horse has over it’s muscles. With inanimate objects like a car, it seems to me that the force applied by...
I’ve been reading about the Kutta Condition and how a vortex results but the information discussing how vortices form aren’t too detailed. I’m just interested to know more about them.
Well, the only two forces on the cart 1 in the horizontal direction are the reaction force applied to it by cart 2 and the force applied by the ground. Thus if cart 1 is accelerating then there is a net force in the forward direction.
Hi there,
Just to map out my question, I am thinking about cart 1 which is a motorized cart and cart 2 -- a cart with a heavy object on it. There is an acceleration rightward.
I understand that Fc1onc2 = -Fc2onc1 which means that the forces do not cancel out as they are applied on different...
It is clear. So I think I used the chain rule correctly in my case. I know I got to the right answer just want to be sure my reasoning was on point and I didn’t misuse the chain rule.
Could I have done the chain rule differently? Just trying to get further concrete understanding.
I forgot where I came across this and why I got so determined to figure it out but I wanted to ask about this d/dx(v^2) business.
My question is to solidify my understanding of the chain rule with physics equations (sorry for crap terminology). Therefore, I know I use it and do the math as...