Friction and Kinematics: Solving for Minimum Time on a Bridge

[RandomGuy1]

Homework Statement

A car starts from rest on half a kilometer long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one cannot drive through the bridge in less than 10 seconds. (Take g = 10 m/s²)

Homework Equations

Frictional force = μN
Kinematic equations

The Attempt at a Solution

I just cannot draw the free body diagram. I took the acceleration to be (a - (mu)g) and it got me nowhere. I get the right answer if I consider ma = μN (I get a = 10m/s² and t = 10s, substituting it in the kinematic equations). But taking ma = μN is like saying the force with which the car is trying to move forward is equal to the force which is pulling it back. But if that is so, how will the car move forward?

 

[Doc Al]

But taking ma = μN is like saying the force with which the car is trying to move forward is equal to the force which is pulling it back. But if that is so, how will the car move forward?

What do you mean by "the force which is pulling it back"? What force is that? Horizontally, there is only a force pushing it forward.

I just cannot draw the free body diagram.

List the forces acting on the car. I see three.

 

[RandomGuy1]

Frictional force is pulling it back. Gravity, Normal Force, Frictional Force are the three forces, right?
If ma is the force with which it is trying to move forward, wouldn't ma being equal to μN mean the car wouldn't move?

 

[Doc Al]

Frictional force is pulling it back.

No, friction is driving it forward!

Gravity, Normal Force, Frictional Force are the three forces, right?

Right.

If ma is the force with which it is trying to move forward, wouldn't ma being equal to μN mean the car wouldn't move?

"ma" is not a force. It's what the net force equals, per Newton's 2nd law: ƩF = ma.

 

[RandomGuy1]

Oh, that makes more sense . Thank you, Doc!