Recent content by Liam C

In my equation they do not have the same sign. In the first line I am adding a positive number (Ff,s) to a negative number (FairR). When I move the FairR to the other side of the equation and subtract, it becomes a double negative, making it a positive. As a result, I add the net force to the...

Fnet = Ff,s + FairR Ff,s = Fnet - FairR Ff,s = 7899.4 Ff,s = 7.90 x 10 ^ 3 N [forward] Thanks again!

Ohh that makes so much more sense! I will think of that from now on.

Thank you for your help, I really appreciate it. FA was supposed to be the applied force from the tire onto the ground and Fr was supposed to be the reaction force from FA. I'm going to try to write an equation with just those three forces now.

I should mention that that diagram is as if the two vehicles were a single object.

Yes, I have already done that, and I have attached it. Based on this I think the sum of the forces equation looks like this: Fnet = Fr + Ff,s - FairR - FA Am I on the right tracK? I am having trouble extracting an answer from this.

Homework Statement A 1450-kg car is towing a trailer of mass 454-kg. The force of air resistance on both vehicles is 7471N[backward]. If the acceleration of both vehicles is 0.225m/s^2[forward], what is the force of static friction on the wheels from the ground? Homework Equations Fnet = ma...

Interesting. Thank you for the help.

Aha I figured it out! I was under the impression that the normal force would be reduced by the extra weight, but in reality it does not change. Us = Ff / FN Us = 7.35 / 49 Us = 0.15 My question now is: why does the force of the normal not change? There is obviously an added upwards force on the...

Okay, I will do that from now on. I think both FN and FA are going up. It doesn't really make sense to subtract one of them from the other when I think about that.

Homework Statement A 5.0kg block of wood sits on a table with a string running over a pulley suspending a mass. The largest mass that can be suspended without moving the block of wood is 0.75kg. a) Calculate the coefficient of friction in this case. b) What type of coefficient of friction has...

Wow that's very interesting, I'll have to bring that up with my teacher. I believe I'm learning "work" next week so I'm excited to find out what that means exactly.

Oh! v2 is likely 7.5m/s then. Aha!

Homework Statement Nikhil is probably eating his daily snack at this time. Old Man Miller comes back from the store and takes the 0.150kg sandwich from Nikhil as he begins to run away. Nikhil is running at 7.5m/s away from me when I hit him in the head with the sandwich that was moving at...

So would the answer to b) be 20N [forward] ? If you separate the system of the ball from the system of the racquet it seems to me that the ball would be propelled forward by the same force it hit the racquet with.