Question

A 1230kg car pushes a 2160kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4590N . Take the friction between the truck's tires and the ground to be 470N .

What is the magnitude F_ConT of the force that the car exerts on the truck? Use the coordinate system shown in Part B, where upward is the positive y direction and to the right, the direction F_ConT points, is the positive x direction.

Answer 1

http://session.masteringphysics.com/problemAsset/1060044/28/1060044FBDa.jpg

Answer 2

i'm on it

Answer 3

the value of F for c on t is 2796N

Answer 4

That answer is incorrect. The value I got was close to that one.

Answer 5

when yu find the accelelration, it solves everything

Answer 6

\[a = \frac{ (F _{ConT}-470) }{ 2160 }, a = \frac{ 4590-F _{TonC} }{ 1230 }\]

Answer 7

\[a=\frac{ F-2f _{f} }{ m _{c} +m _{t}}\]
\[F-f _{f}-f _{t on c} = m _{c}a\]
\[f _{c on t} - f _{f} = m _{t}a\]

Answer 8

so if if yu add the two equations, the f t on c cancels with the f c on t, cause they are the same just different direction

Answer 9

What is capital F in your equation.

Answer 10

I think your setup may be correct but the two friction forces are different in this environment

Answer 11

the F is the N
no then we will have to use \[muN\] as the friction force

Answer 12

Got the answer, thank you.

Answer 13

which did u use?
which formula?

Answer 14

\[a = \frac{f _{C}-f _{T} }{ M _{C}+M _{T}} \rightarrow M _{T}(a)+f _{T}=F _{ConT}\]